The extent to which any acid gives off protons is the extent to which it is ionized, and this is a function of a property of the acid known as its Ka, which you can find in Therefore, you simply use the molarity of the solution provided for [HA], which in this case is 0.10. Step 3: Calculate the Percent Ionization. Percent Ionization Formula Percent ionization can be computed by dividing the concentration of ionized acid or base in equilibrium by the original Hence, don’t hesitate to contact us to share your views about How To Calculate Percent Ionization , we will listen carefully and try to improve our site.

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## Percent Calculators | Calculates Percentages & Explains How

These Percent Calculators find answers to common problems on percentages. Get simple percentages or calculate the percent of The percent of increase or decrease is the measure of percent change.Percentage calculator is an online tool to calculate percentages. Percentage calculator to find percentage of a number, calculate x as a For example, 12% (read as ‘twelve percent’) is equal to 12/100, or 3/25, or 0.12.Table calculations, as discussed in Using Table Calculations, make it easy to create on-the-fly metrics and enable the creation of This is great for calculating metrics like percent of previous and percent change.To calculate 25 percent of a value, multiply the value by 0.25. There is really no way to calculate nickel percent in chocolates. There will be a different amount in all kinds of chocolates and none in some.

Calculation of percentage error. Why calculate percentages and rectify errors? Standard symbol of percentage (Photo Credit: SimonBrun/ Shutterstock). According to the Cambridge dictionary, the term percentage means “an amount of something, often expressed as a number out of 100.The extent to which any acid gives off protons is the extent to which it is ionized, and this is a function of a property of the acid known as its Ka, which you can find in Therefore, you simply use the molarity of the solution provided for [HA], which in this case is 0.10. Step 3: Calculate the Percent Ionization.Percent Ionization, like equilibrium expressions, is a way for chemists to quantify the extent of a reaction. Mr.Use our free calculator to calculate the percent change between two numbers.

## Percentage Calculator

Calculate percent value,percent change,percentage increase and decrease and compound Use Alcula’s percentage calculator to compute percentages and answer questions such as 250 is 8 percent of what amount? How much is 12000+8%.Related Threads on Chemistry problem, calculate percent ionization. Insights How Quantum Information Theorists Revealed the Relativity Principle at the Foundation of Quantum Mechanics.This free percentage calculator computes a number of values involving percentages, including the percentage difference between two given values. Explore various other math calculators as well as hundreds of calculators addressing finance, health, fitness and more.Calculate percent, percent change, percentage points, and percentile. Percent Change. By how much did the thing adjust? Percentage Points.

Express the percent ionization to two significant digits. We review their content and use your feedback to keep the quality high.Calculating percent is a fundamental math skill. Here’s a step-by-step tutorial on how to calculate percent. Percentages are used to make car and house payments, calculate tips and pay taxes on goods. Percent calculations are fundamental to many classes, especially science courses.But first, let’s define how to calculate percentage mathematically. 2. Mathematical Formula. In mathematics, a percentage is a number or ratio expressed as a fraction of 100. It’s often denoted using the percent sign, “%”. Let’s consider a student that obtains x marks out of total y marks.The term annual percentage rate of charge (APR), corresponding sometimes to a nominal APR and sometimes to an effective APR (EAPR), is the interest rate for a whole year (annualized), rather than just a monthly fee/rate, as applied on a loan, mortgage loan, credit card, etc.

## How To Calculate Percent Ionization – 07/2021

Percent Ionization Formula Percent ionization can be computed by dividing the concentration of ionized acid or base in equilibrium by the original Hence, don’t hesitate to contact us to share your views about How To Calculate Percent Ionization , we will listen carefully and try to improve our site.This solution provides a step-by-step explanation of how to calculate the percent ionization for the weak acid in the following problem. The required formulas and necessary stoichiometry for the problem are included within the solution. To view the solution, a word file attachment must be opened.Similar Questions. Chemistry. Calculate the percent ionization of cyanic acid, Ka=2.0×10^-4, in a buffer soln that is .50M HCNO and .10M NaCNO. What will happen if a small amount of hydrochloric acid is added to a 0.1 M solution of HF? A) The percent ionization of HF will increase.Percent Ionization – . instead of characterizing a weak acid by its k a , we can calculate how much it ionizes. the. Presentation Transcript. [H3O+]eq [HA]initial Calculating Percent Ionization • Percent Ionization = 100 • In this example [H3O+]eq = 4.2 10−3 M [HCOOH]initial = 0.10 M.

In order to calculate percentages online, click the button. The result appears on the next page. Our purpose is to offer the best online percentage calculator with fast and useful answers. First we check how much is one percent: we divide 900 by 100. We get 9.Percent change calculation is very easy, if you understand the basics of the calculation method and know how to apply the formula. You can calculate year to year change to understand the rate of inflation or deflation from one period to the next.How to calculate percentage discount With This Percent off / Discount Calcul. ator: First of all, you have to convert the 15% discount to decimal, to do so, just divide 15 by 100 i:e 0.15; Then, simply multiply 0.Calculate the pH of a Weak Acid and Percent Ionization. • How to Use Henderson Hasselbalch Equation for Pharmacology.

## calculating the concentrations using the percent ionization

A 0.100 M solution of bromoacetic acid, is 13.2 percent ionized. Calculate [BrCH2COO], [H], [BrCH2COOH], and the Ka for Yes you need an ICE table. If you start with 0.100M and it becomes 13.2% ionised you will end up with 0.0132M as [BrCH2COO] and, [H] and be left with 0.100-0.A calibration curve of Pb was calculated to have the equaiton of y=0.01 x +0.003. y=absorbance and x=concentration of Pb. The concentrations in raw and spiked sample were found using the formula as 5.6 ppm and 6.1 ppm respectively. (This is before considering the DF).Percent Daily Value Explained The percent Daily Value (%DV) shows how much a nutrient in a serving of food contributes to a total daily diet. For example, the Daily Value for saturated fat is less than 20 grams (g) per day (based on a 2,000 calorie daily diet), which equals 100% DV.Percent ionization is the quantity of a weak acid that ionizes in a solution expressed as a percentage. The percent ionization of a given acid varies depending on the original concentration of the acid. Formula to calculate percent ionization.

How to calculate percentage? answered in detail. Examples of using a percentage calculator.1) Calculate the percent dissociation of HA in a 0.10 M solution. Problem #3: Calculate the degree of ionization of acetic acid in the following solutions Comment: note how the solution was diluted from 5 mL to 10 mL, thus cutting the concentration in half.From month-to-month, to year-over-year, to annualized here are the formulas and concepts you need to know to understand percent change. Where it gets complicated is how we talk about this arithmetic.To understand theta, it is important to first understand the difference between the intrinsic and extrinsic value of an option. It can be used to see how much value an option loses on a daily basis, and how much the underlying asset will need to change to offset the loss in according to theta.

## How To Find PH, POH, H3O+, And OH- STEP BY STEP

We'll go over all the formulas you needto find ph, poh, acid concentration and base concentration. Hello everyone i'mmelissa maribel your personal tutor and let's go over two potential testquestions. In question 1, we are given the concentration of acid and asked to findthe ph, poh and base concentration to find the ph we will use the ph formulaand plug in the given acid concentration. Alright,important significant figure note, ph and poh have specific significant figurerules when you are given an acid or base concentration the amount of sig figs youhave will be the amount of decimal places the ph and poh should be roundedto, this is why we get 3. 39 as our ph and note our poh will also have two decimalplaces. Since we just found the ph we can use this formula and plug in ph to findpoh subtract both sides by three point three nineand our poh is ten point six one since we just found poh and we are now findingthe concentration of base we can use this formula and plug in the poh. Plugging this into your calculator make sure to press 2nd log and then put theexponent of negative ten point six one. Round to two sig figs since our initialacid concentration had only two sig figs.

this gives us two point five times tento the negative 11th and our units are in molarity since concentration ismeasured in molarity. In question two we are given the ph and asked to find ouracid and base concentrations and the poh starting with part a since we are giventhe ph we can use this formula and solve for that acid concentration so press 2ndlog on your calculator then input the exponent of negative 8. 5 and we get thisas our acid concentration. Significant figure note when we're going from ph toacid concentration the number of decimal places for the phtells you how many sig figs to round to this is why our acid concentration onlyhas one sig fig. Part b to find our base concentration we will use this formulaand plug in the acid concentration we've just found divide both sides and we willget this as our base concentration rounded to one sig fig. Part c there aretwo formulas we can use to find poh in this case here's the first formula we'llplug in the base concentration to get this as our poh the second formula wecould have used instead was this one and we can plug in our ph subtract bothsides by 8. 5 and we would get the same value.

now here's a summary of when touse each formula when going from your acid concentration to ph use thisformula. Going the opposite way from ph to acid concentration use this. Now whenyou're going from your base concentration to poh use this formula. Going the opposite way from poh to base concentration use this if you are giventhe ph and asked to find poh or going from poh to ph use this formula. Goingfrom acid concentration to your base concentration or vice versa baseconcentration to acid concentration use this.

if you want to cut your study timein half i created detailed notes on everything you need to know on acids andbases you can find that in the description box below and make sure tocheck out this other helpful video and remember stay determined you can do this! . We'll go over all the formulas you needto find ph, poh, acid concentration and base concentration. Hello everyone i'mmelissa maribel your personal tutor and let's go over two potential testquestions. In question 1, we are given the concentration of acid and asked to findthe ph, poh and base concentration to find the ph we will use the ph formulaand plug in the given acid concentration. Alright,important significant figure note, ph and poh have specific significant figurerules when you are given an acid or base concentration the amount of sig figs youhave will be the amount of decimal places the ph and poh should be roundedto, this is why we get 3. 39 as our ph and note our poh will also have two decimalplaces. Since we just found the ph we can use this formula and plug in ph to findpoh subtract both sides by three point three nineand our poh is ten point six one since we just found poh and we are now findingthe concentration of base we can use this formula and plug in the poh. Plugging this into your calculator make sure to press 2nd log and then put theexponent of negative ten point six one. Round to two sig figs since our initialacid concentration had only two sig figs.

this gives us two point five times tento the negative 11th and our units are in molarity since concentration ismeasured in molarity. In question two we are given the ph and asked to find ouracid and base concentrations and the poh starting with part a since we are giventhe ph we can use this formula and solve for that acid concentration so press 2ndlog on your calculator then input the exponent of negative 8. 5 and we get thisas our acid concentration. Significant figure note when we're going from ph toacid concentration the number of decimal places for the phtells you how many sig figs to round to this is why our acid concentration onlyhas one sig fig. Part b to find our base concentration we will use this formulaand plug in the acid concentration we've just found divide both sides and we willget this as our base concentration rounded to one sig fig. Part c there aretwo formulas we can use to find poh in this case here's the first formula we'llplug in the base concentration to get this as our poh the second formula wecould have used instead was this one and we can plug in our ph subtract bothsides by 8. 5 and we would get the same value.

now here's a summary of when touse each formula when going from your acid concentration to ph use thisformula. Going the opposite way from ph to acid concentration use this. Now whenyou're going from your base concentration to poh use this formula. Going the opposite way from poh to base concentration use this if you are giventhe ph and asked to find poh or going from poh to ph use this formula. Goingfrom acid concentration to your base concentration or vice versa baseconcentration to acid concentration use this.

if you want to cut your study timein half i created detailed notes on everything you need to know on acids andbases you can find that in the description box below and make sure tocheck out this other helpful video and remember stay determined you can do this! .

## Mass Percent Of A Solution Made Easy: How To Calculate Mass % Or Make A Specific Concentration

Welcome to mass percent made easy, broughtto you by ketzbook in this video, we are going to learn how tocalculate the mass percent of a solution and how to calculate the mass of the solute orsolvent if you know the mass percent concentration. But first, what is mass percent? Mass percent is a way to measure concentration. “per” means for each, and “cent” islatin for 100, so percent refers to the amount of solute dissolved in 100 units of solution. “mass” simply refers to how we measureboth the solute and the solution. For example, the fat content in milk is measuredin mass percent. What is 2% milk? It means that 100 grams of that milk contains2 grams of fat.

there are two grams of fat for each 100 gramsof 2% milk. The equation that we will use to calculatemass percent is: mass percent equals the mass of the solute divided by the total mass of the solution all multiplied by 100 to turn the fraction into a percent. Let’s try some examples to see how thisworks. Calculate the mass percent concentration of a 38. 5 gram aqueous solution that contains 1. 96 grams of sodium bromide. Before we solve this problem, let’s makesome sense of it.

first, the term aqueous means that water isthe solvent. Next, the problem tells us that we are solvingfor the mass percent. 38. 5 grams is the mass of the solution, and1. 96 grams is the mass of the solute. Remember that the solute is the thing thatis dissolved in the solvent. Now we’re ready to plug in some numbersto calculate the mass percent.

1. 96 grams goes on the top of the fractionand 38. 5 grams goes on the bottom of the fraction. This fraction is then multiplied by 100%. In your calculator, type 1. 96 divided by 38. 5times 100, which gives us the answer of 5. 09 percent. As for the units, notice that grams on thetop and bottom of the fraction cancel each other out. This is true for any kind of percent or anykind of parts per calculation.

the units on the top and bottom of the fractionmust be the same so that they cancel out. Percent functions as the units of our answer,but in reality percent is merely a way to represent a dimensionless ratio. Okay, let’s try another problem. What is the mass percent of 125 grams of fructosedissolved in 375 ml of water? Once again, we are solving for the mass percentconcentration, 125 grams is the mass of the solute, and 375 ml is the volume of the solvent. Volume? What do we do with volume? Water is the solvent, so this is relativelysimple because the density of water is approximately 1 g/ml at room temperature.

so for water, 1 ml has a mass of 1 gram. That means that 375 ml is the same as 375grams. As long as we don’t need any more than 3significant figures and as long as the water is not too warm, this approximation worksfine. If we did need more precision, we would haveto calculate the mass of the solvent based on its density at the given temperature. Now we can plug in some numbers.

the mass percent is equal to the mass of thesolute divided by the mass of the solution all times 100 percent. We put 125 grams in for the mass of the solute. You may be tempted to put 375 grams in thebottom, but that is only the mass of the solvent. The mass of the solution is the mass of thesolvent plus the mass of the solute, that is 375 plus 125. That works out to 125 divided by 500 times100, or 25%.

suppose, however, we wanted to solve for somethingdifferent, such as the mass of the solute. Let’s try this next problem. How much sodium fluoride is in a 221 g tubeof toothpaste that has 0. 24% sodium fluoride? In this case, we are solving for the massof the solute. 221 grams is the mass of the solution, and 0. 24% is the mass percent concentration of sodium fluoride. We can set the problem up the same as always.

the mass percent is 0. 24%, which is equalto the mass of the solute divided by the mass of the solution or 221 grams, all multipliedby 100%. The only difference this time is that what we are solving for is on the right side of the equation, so we will need to do a bit of rearranging. First, let’s remove the units because percentis on both sides of the equation, and we know that the mass of the solute has the units of grams. Next, we can multiply both sides of the equationby 221, which cancels it out on the right. Finally, we can divide both sides of the equationby 100 to cancel it out on the right side as well.

this leaves us with the mass of the solutebeing equal to 221 times 0. 24 divided by 100, which calculates to be 0. 53 grams. That is approximately one tenth the lethaldose of sodium fluoride for adults, but it is enough to be dangerous for small children,so caution should be exercised when allowing children to use any fluoride containing product. Okay, let’s try one last problem. How could you make a 7. 8% aqueous solutionof glucose using 5. 0 g of glucose? Like always, let’s first identify the knownsand the unknowns in the problem. 7. 8% is the mass percent concentration ofthe solution we want to make, and 5 grams is the mass of the solute we need to use.

the question doesn’t specifically say what we are solving for, but we can infer it from what is missing. Mass percent equals the mass of the solute divided by the mass of the solution all times 100 percent. The mass percent is 7. 8, and the mass of thesolute is 5. That leaves the mass of the solution as theonly unknown, so we should solve for that. However, when you make a solution, it is morepractical to weigh the solvent all by itself, so we should ultimately solve for the massof the solvent.

in order to solve this problem, we need todo a little bit of rearranging. The variable we are solving for cannot bein the denominator, so the first thing we need to do is multiply both sides of the equationby the mass of the solution. Mass of the solution then cancels out on theright side of the equation. Next, we need to get the mass of the solutionall by itself, so we divide both sides by 7. 8. The 7. 8 on the top and bottom of the fractioncancel each other out.

this leaves us with the mass of the solution equals 5 times 100 divided by 7. 8, which calculates to be 64. 1. Because the units for the mass of glucose are grams, the units for the mass of solution are also grams. We are almost done, but as we mentioned earlier,it is easier in the lab to weigh the solvent separately, so we need to calculate the massof the solvent all by itself. The mass of the solution equals the mass ofthe solvent plus the mass of the solute, and we know that the mass of the solute is 5 grams. In other words, the mass of the solvent is64. 1 minus 5 or 59. 1 grams.

by the way, the reason i have included anextra significant figure for this quantity is because it is not based on a measurementbut something we will make in the lab, so we want it to be as precise as possible. Now that we have the numbers we need, we cango ahead and weigh everything we need. Weigh the glucose in a weighing boat or aclean weighing container. Although we could use a graduated cylinder to measure the water, it is better to weigh the water. Once we have the correct amounts, simply combinethem in a beaker or flask, and stir it a little until it is all dissolved.

the resulting solution is a 7. 8 percent aqueousglucose solution. Thanks for watching. If you found this video useful, please likeor subscribe. Feel free to share any comments or questionsyou have below, and check me out at ketzbook. Com. .

welcome to mass percent made easy, broughtto you by ketzbook in this video, we are going to learn how tocalculate the mass percent of a solution and how to calculate the mass of the solute orsolvent if you know the mass percent concentration. But first, what is mass percent? Mass percent is a way to measure concentration. “per” means for each, and “cent” islatin for 100, so percent refers to the amount of solute dissolved in 100 units of solution. “mass” simply refers to how we measureboth the solute and the solution. For example, the fat content in milk is measuredin mass percent. What is 2% milk? It means that 100 grams of that milk contains2 grams of fat.

there are two grams of fat for each 100 gramsof 2% milk. The equation that we will use to calculatemass percent is: mass percent equals the mass of the solute divided by the total mass of the solution all multiplied by 100 to turn the fraction into a percent. Let’s try some examples to see how thisworks. Calculate the mass percent concentration of a 38. 5 gram aqueous solution that contains 1. 96 grams of sodium bromide. Before we solve this problem, let’s makesome sense of it.

first, the term aqueous means that water isthe solvent. Next, the problem tells us that we are solvingfor the mass percent. 38. 5 grams is the mass of the solution, and1. 96 grams is the mass of the solute. Remember that the solute is the thing thatis dissolved in the solvent. Now we’re ready to plug in some numbersto calculate the mass percent.

1. 96 grams goes on the top of the fractionand 38. 5 grams goes on the bottom of the fraction. This fraction is then multiplied by 100%. In your calculator, type 1. 96 divided by 38. 5times 100, which gives us the answer of 5. 09 percent. As for the units, notice that grams on thetop and bottom of the fraction cancel each other out. This is true for any kind of percent or anykind of parts per calculation.

the units on the top and bottom of the fractionmust be the same so that they cancel out. Percent functions as the units of our answer,but in reality percent is merely a way to represent a dimensionless ratio. Okay, let’s try another problem. What is the mass percent of 125 grams of fructosedissolved in 375 ml of water? Once again, we are solving for the mass percentconcentration, 125 grams is the mass of the solute, and 375 ml is the volume of the solvent. Volume? What do we do with volume? Water is the solvent, so this is relativelysimple because the density of water is approximately 1 g/ml at room temperature.

so for water, 1 ml has a mass of 1 gram. That means that 375 ml is the same as 375grams. As long as we don’t need any more than 3significant figures and as long as the water is not too warm, this approximation worksfine. If we did need more precision, we would haveto calculate the mass of the solvent based on its density at the given temperature. Now we can plug in some numbers.

the mass percent is equal to the mass of thesolute divided by the mass of the solution all times 100 percent. We put 125 grams in for the mass of the solute. You may be tempted to put 375 grams in thebottom, but that is only the mass of the solvent. The mass of the solution is the mass of thesolvent plus the mass of the solute, that is 375 plus 125. That works out to 125 divided by 500 times100, or 25%.

suppose, however, we wanted to solve for somethingdifferent, such as the mass of the solute. Let’s try this next problem. How much sodium fluoride is in a 221 g tubeof toothpaste that has 0. 24% sodium fluoride? In this case, we are solving for the massof the solute. 221 grams is the mass of the solution, and 0. 24% is the mass percent concentration of sodium fluoride. We can set the problem up the same as always.

the mass percent is 0. 24%, which is equalto the mass of the solute divided by the mass of the solution or 221 grams, all multipliedby 100%. The only difference this time is that what we are solving for is on the right side of the equation, so we will need to do a bit of rearranging. First, let’s remove the units because percentis on both sides of the equation, and we know that the mass of the solute has the units of grams. Next, we can multiply both sides of the equationby 221, which cancels it out on the right. Finally, we can divide both sides of the equationby 100 to cancel it out on the right side as well.

this leaves us with the mass of the solutebeing equal to 221 times 0. 24 divided by 100, which calculates to be 0. 53 grams. That is approximately one tenth the lethaldose of sodium fluoride for adults, but it is enough to be dangerous for small children,so caution should be exercised when allowing children to use any fluoride containing product. Okay, let’s try one last problem. How could you make a 7. 8% aqueous solutionof glucose using 5. 0 g of glucose? Like always, let’s first identify the knownsand the unknowns in the problem. 7. 8% is the mass percent concentration ofthe solution we want to make, and 5 grams is the mass of the solute we need to use.

the question doesn’t specifically say what we are solving for, but we can infer it from what is missing. Mass percent equals the mass of the solute divided by the mass of the solution all times 100 percent. The mass percent is 7. 8, and the mass of thesolute is 5. That leaves the mass of the solution as theonly unknown, so we should solve for that. However, when you make a solution, it is morepractical to weigh the solvent all by itself, so we should ultimately solve for the massof the solvent.

in order to solve this problem, we need todo a little bit of rearranging. The variable we are solving for cannot bein the denominator, so the first thing we need to do is multiply both sides of the equationby the mass of the solution. Mass of the solution then cancels out on theright side of the equation. Next, we need to get the mass of the solutionall by itself, so we divide both sides by 7. 8. The 7. 8 on the top and bottom of the fractioncancel each other out.

this leaves us with the mass of the solution equals 5 times 100 divided by 7. 8, which calculates to be 64. 1. Because the units for the mass of glucose are grams, the units for the mass of solution are also grams. We are almost done, but as we mentioned earlier,it is easier in the lab to weigh the solvent separately, so we need to calculate the massof the solvent all by itself. The mass of the solution equals the mass ofthe solvent plus the mass of the solute, and we know that the mass of the solute is 5 grams. In other words, the mass of the solvent is64. 1 minus 5 or 59. 1 grams.

by the way, the reason i have included anextra significant figure for this quantity is because it is not based on a measurementbut something we will make in the lab, so we want it to be as precise as possible. Now that we have the numbers we need, we cango ahead and weigh everything we need. Weigh the glucose in a weighing boat or aclean weighing container. Although we could use a graduated cylinder to measure the water, it is better to weigh the water. Once we have the correct amounts, simply combinethem in a beaker or flask, and stir it a little until it is all dissolved.

the resulting solution is a 7. 8 percent aqueousglucose solution. Thanks for watching. If you found this video useful, please likeor subscribe. Feel free to share any comments or questionsyou have below, and check me out at ketzbook. Com. .

## Isotopes, Percent Abundance, Atomic Mass | How To Pass Chemistry

In this video you're finally gonnaunderstand what an isotope is and how to calculate percent abundance. All rightlet's do this. Hello hello melissa maribel hereand i help students like you understand what you just learned in class so youstress less and you graduate faster. Before we get started let's take a deepbreath. It's gonna be a lot easier than you think. Starting with our first example let's just talk about the concept of isotopes. So back in 2009 i weighed a hundred and twenty pounds. Then in 2012 i weighed ahundred and thirty pounds and then in 2015 when i discovered my love for sushii weighed a hundred and forty pounds.

so even though there were different weightsor different versions of myself, i am still the same person. Isotopes aredifferent versions of the same element. So how does this relate back tochemistry. Let's talk about gold. The chemical symbol for gold is au.

takethis gold bar, it weighs 196 amu. Our second gold bar weighs 197 amu and ourthird gold bar weighs 198 amu. All of these gold bars have different weightsbut are made out of the same element which is gold. These are three differenttypes of isotopes of gold. If we were to look at the periodic table and look forwhere au is or gold, we'd see it has a bottom number, right? That atomic mass. That atomic mass is actually the average of all the different types of isotopesof that element.

isotopes have the same atomic number, the same protons andelectrons. However, they do not have the same masses. So they have different masses and a different number of neutrons. Let's gointo our first example talking about percent abundance. A certain element xhas four isotopes 0. 5600% of x has a mass of 83. 91343 amu. Second isotope, 9. 860% of x has a mass of 85. 90927 amu.

ourthird isotope, 7. 000% of x has a mass of 86. 90890 amu and our last isotope 82. 58% of x has a mass of 87. 90562 amu. Find the average atomic mass of element x. You have four differentisotopes, those are our givens. That 0. 5600% of our certainamount of mass. So this is what we refer to as percent abundance, where we have acertain amount of percent of our first isotope of a mass.

so you see it changesfor every single isotope, however, if we were to add up all the percentages, it'salways out of a hundred percent. In this example, you are finding your averageatomic mass. Step one is to convert all percentages to decimals by dividing by 100. Your first given 0. 5600. We'll divide that by 100. And you get 0. 005600 there's a trick for this. What we can instead do is move the decimal place over twice to the left for every single percentage to get ourdecimal form and we'll see we keep doing this and it gives us our answer.

movingon we'll jump to our actual percent abundance formula where our atomic massis equal to the percentage converted to a decimal, times, the mass of thatspecific isotope and we'll keep adding every different type of isotope. Since wehave four isotopes, i went ahead and added our first decimalform of that percentage multiplied by the mass of that first isotope and keptdoing that for every single isotope since we have four. Looking at this firstportion we'll multiply 0. 005600, times the massof 83. 91343 and that gives us 0. 469915. We would then do this for every single isotopeand you'll get this. We'll add them all together,that'll give us 87. 616626 amu.

the reason whywe're going to round to only four sig figs is because going back to our givenwe actually only had four significant figures, 82. 58 isfour sig figs, same with 7. 000 and you'll see that everysingle one actually has four sig figs. So we'll round up and our final atomic massis 87. 62 amu. In the second example we are solving forpercent composition or percent abundance we're no longer solving for the averageatomic mass. Naturally occurring copper consists of cu at 63, this is just theisotope notation, where our mass is equal to 62. 9296 amu and our second isotope, copper at 65, which our mass is 64. 9278 amu, quick note, they're just rounding here. This is the mass numberbut they do give us the exact mass amount.

continuing with this question,with an average atomic mass of 63. 546 amu. What isthe percent abundance of copper in terms of these two isotopes? You're given, arethe two different types of isotopes so we're really just focusing on our massesthat are exact in this case. This is just the proper notation that they typicallyhave it, but as mentioned we're just focusing on these masses. We're alsogiven the average atomic mass of 63. 546 amu and we'reasked to find the percent composition of copper. What they're actually referringto here is, they're saying how much of this isotope, what percentage, is withinour atomic mass? So we're really looking for two different percentages for ourtwo different isotopes.

using our percent abundance formula once more, our atomicmass is equal to that percentage however we don't know what the percentage is ofour first isotope. So in that case we're actually just going to refer to this asx. I also went ahead and placed what we do know which was that atomic mass. Soreferring to this percentage as x, it's just a way to be able to figure out whatx is towards the end. Our mass of our first isotope was this.

and next we onceagain do not know what the second percentage of our isotope is. So whatwe'll do is we'll subtract it from 1. What 1 refers to is really saying, it'sout of a hundred percent so if we don't know what x is, we'll then subtract itfrom 1 or a hundred percent and that would then give us our second percentageof our isotope. We'll multiply this by our mass of our second isotope. Next,going back to math, let's go ahead and actually distributethis mass to both the one and that negative x.

when we do this we end upgetting this, and you want to group your x's together so we'll combine these twox values. We're actually subtracting since this is a negative 64. 9278. When we do that we end up getting negative 1. 9982x. From there we want to isolate our x value. So we'll subtractover that 64. 9278 to both sides. These would then cancel and subtracting these two we get a negative 1. 3818. And we want to isolate that x by itself, so we'll getrid of that negative 1. 9982 by dividing it for both sides. Those would then cancel and our x value gives us a positive value of 0. 6915.

that was just for our first isotope which was that copper at63. For our second isotope what we'll do is, taking that 1 minus x,we're actually going to subtract this x value from the 1 and that would thengive us 0. 30847. The next step is to multiply bothof these decimals by 100. The reason is because we want this to be apercentage, so multiplying this by 100 turns it into our percentages. We'll see that our first isotope, that copper 63, has a percentage of 69. 15%. Our second isotope, where copper 65, has a percentage of30. 85%.

these are our percent compositions of ourisotopes. Now that we've worked on two possible exam questions,hint hint, it's your turn to see if you're ready for your exam. Musica maestro! So are you ready for your exam? As atutor, i always tell my students to do these practice problems, to get them down,because i do see them on exams over and over again. Teachers like to reuse theirexam questions, so if you feel like you're not ready, you need some more help,check the description box below and you can find my available times for tutoring. And make sure to like, subscribe and i'll see you guys next time. .

in this video you're finally gonnaunderstand what an isotope is and how to calculate percent abundance. All rightlet's do this. Hello hello melissa maribel hereand i help students like you understand what you just learned in class so youstress less and you graduate faster. Before we get started let's take a deepbreath. It's gonna be a lot easier than you think. Starting with our first example let's just talk about the concept of isotopes. So back in 2009 i weighed a hundred and twenty pounds. Then in 2012 i weighed ahundred and thirty pounds and then in 2015 when i discovered my love for sushii weighed a hundred and forty pounds.

so even though there were different weightsor different versions of myself, i am still the same person. Isotopes aredifferent versions of the same element. So how does this relate back tochemistry. Let's talk about gold. The chemical symbol for gold is au.

takethis gold bar, it weighs 196 amu. Our second gold bar weighs 197 amu and ourthird gold bar weighs 198 amu. All of these gold bars have different weightsbut are made out of the same element which is gold. These are three differenttypes of isotopes of gold. If we were to look at the periodic table and look forwhere au is or gold, we'd see it has a bottom number, right? That atomic mass. That atomic mass is actually the average of all the different types of isotopesof that element.

isotopes have the same atomic number, the same protons andelectrons. However, they do not have the same masses. So they have different masses and a different number of neutrons. Let's gointo our first example talking about percent abundance. A certain element xhas four isotopes 0. 5600% of x has a mass of 83. 91343 amu. Second isotope, 9. 860% of x has a mass of 85. 90927 amu.

ourthird isotope, 7. 000% of x has a mass of 86. 90890 amu and our last isotope 82. 58% of x has a mass of 87. 90562 amu. Find the average atomic mass of element x. You have four differentisotopes, those are our givens. That 0. 5600% of our certainamount of mass. So this is what we refer to as percent abundance, where we have acertain amount of percent of our first isotope of a mass.

so you see it changesfor every single isotope, however, if we were to add up all the percentages, it'salways out of a hundred percent. In this example, you are finding your averageatomic mass. Step one is to convert all percentages to decimals by dividing by 100. Your first given 0. 5600. We'll divide that by 100. And you get 0. 005600 there's a trick for this. What we can instead do is move the decimal place over twice to the left for every single percentage to get ourdecimal form and we'll see we keep doing this and it gives us our answer.

movingon we'll jump to our actual percent abundance formula where our atomic massis equal to the percentage converted to a decimal, times, the mass of thatspecific isotope and we'll keep adding every different type of isotope. Since wehave four isotopes, i went ahead and added our first decimalform of that percentage multiplied by the mass of that first isotope and keptdoing that for every single isotope since we have four. Looking at this firstportion we'll multiply 0. 005600, times the massof 83. 91343 and that gives us 0. 469915. We would then do this for every single isotopeand you'll get this. We'll add them all together,that'll give us 87. 616626 amu.

the reason whywe're going to round to only four sig figs is because going back to our givenwe actually only had four significant figures, 82. 58 isfour sig figs, same with 7. 000 and you'll see that everysingle one actually has four sig figs. So we'll round up and our final atomic massis 87. 62 amu. In the second example we are solving forpercent composition or percent abundance we're no longer solving for the averageatomic mass. Naturally occurring copper consists of cu at 63, this is just theisotope notation, where our mass is equal to 62. 9296 amu and our second isotope, copper at 65, which our mass is 64. 9278 amu, quick note, they're just rounding here. This is the mass numberbut they do give us the exact mass amount.

continuing with this question,with an average atomic mass of 63. 546 amu. What isthe percent abundance of copper in terms of these two isotopes? You're given, arethe two different types of isotopes so we're really just focusing on our massesthat are exact in this case. This is just the proper notation that they typicallyhave it, but as mentioned we're just focusing on these masses. We're alsogiven the average atomic mass of 63. 546 amu and we'reasked to find the percent composition of copper. What they're actually referringto here is, they're saying how much of this isotope, what percentage, is withinour atomic mass? So we're really looking for two different percentages for ourtwo different isotopes.

using our percent abundance formula once more, our atomicmass is equal to that percentage however we don't know what the percentage is ofour first isotope. So in that case we're actually just going to refer to this asx. I also went ahead and placed what we do know which was that atomic mass. Soreferring to this percentage as x, it's just a way to be able to figure out whatx is towards the end. Our mass of our first isotope was this.

and next we onceagain do not know what the second percentage of our isotope is. So whatwe'll do is we'll subtract it from 1. What 1 refers to is really saying, it'sout of a hundred percent so if we don't know what x is, we'll then subtract itfrom 1 or a hundred percent and that would then give us our second percentageof our isotope. We'll multiply this by our mass of our second isotope. Next,going back to math, let's go ahead and actually distributethis mass to both the one and that negative x.

when we do this we end upgetting this, and you want to group your x's together so we'll combine these twox values. We're actually subtracting since this is a negative 64. 9278. When we do that we end up getting negative 1. 9982x. From there we want to isolate our x value. So we'll subtractover that 64. 9278 to both sides. These would then cancel and subtracting these two we get a negative 1. 3818. And we want to isolate that x by itself, so we'll getrid of that negative 1. 9982 by dividing it for both sides. Those would then cancel and our x value gives us a positive value of 0. 6915.

that was just for our first isotope which was that copper at63. For our second isotope what we'll do is, taking that 1 minus x,we're actually going to subtract this x value from the 1 and that would thengive us 0. 30847. The next step is to multiply bothof these decimals by 100. The reason is because we want this to be apercentage, so multiplying this by 100 turns it into our percentages. We'll see that our first isotope, that copper 63, has a percentage of 69. 15%. Our second isotope, where copper 65, has a percentage of30. 85%.

these are our percent compositions of ourisotopes. Now that we've worked on two possible exam questions,hint hint, it's your turn to see if you're ready for your exam. Musica maestro! So are you ready for your exam? As atutor, i always tell my students to do these practice problems, to get them down,because i do see them on exams over and over again. Teachers like to reuse theirexam questions, so if you feel like you're not ready, you need some more help,check the description box below and you can find my available times for tutoring. And make sure to like, subscribe and i'll see you guys next time. .