mole practice

7/28/2019 Mole Practice

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The Mole Concept

1. Relative Atomic Masses

John Dalton was the first chemist to compile a list of relative atomic masses for atoms. At that time no one

could weigh individual atoms so he did experiments which resulted in determining the relative masses of atoms. Since hydrogen is the lightest element, he arbitrarily assigned the mass of a hydrogen atom to be 1 amu(atomic mass unit). By reacting hydrogen with other elements he could determine their mass relative tohydrogen.

From his experiments Dalton found that an oxygen atom has a relative mass of 16 amu. This means that anoxygen atom is 16 times as heavy as a hydrogen atom.

Since Dalton, we have used elements other than hydrogen as the standard. Oxygen was once used and carbon isnow the standard for comparison. But in any event a value of 1 amu for hydrogen is still the origin of our

present day tables of relative atomic masses.

2. The Mole

Amadeos Avogadro had an inspiration that greatly simplified calculations in chemistry. He reasoned that, if anatom of hydrogen weighed 1 amu and an atom of oxygen weighed 16 amu, by weighing out 1 gram of hydrogenand 16 grams of oxygen we would have an equal number of atoms. In fact if we weighed out the relativeatomic mass of any element in grams we would always have the same number of atoms. He called that number of atoms a mole .

Allthe Same

Number of Atoms

One Mole

1 g H 16 g O

12 g C 24 gMg


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The word mole is similar to other words in our language which stand for a particular count. When you buydonuts you often ask for a dozen donuts. Whether you buy a dozen chocolate donuts or a dozen glazed donutsor a dozen jelly donuts --- a dozen donuts is always the same number of donuts.

Likewise, one mole of oxygen atoms or one mole of carbon atomsor one mole of hydrogen atoms is always the same number of atoms. Avogadro was not able to determine how many atoms werein a mole, but that didnt matter. All that mattered was that a moleof any kind of atom was the same number of atoms.

Today we are able to do experiments to allow us to count atoms.We now know that one mole of atoms equals 6.02 x 10 23 atoms.

dozen 12gross 144 pair 2mole 6.02 x 10 23

1 mole of hydrogen atoms = 6.02 x 10 23 hydrogen atoms1 mole of donuts = 6.02 x 10 23 donuts

1 mole of pencils = 6.02 x 1023

pencils

3. Molar Mass

We have already learned that the mass of one mole of any kind of atom can be determined by looking up thatelement on a list of relative atomic masses and expressing that mass in grams.

This same concept may be applied to molecules. Just add up the masses of the atoms to get the mass of onemole of a particular kind of molecule. Look over the following examples to see how to determine the molar mass of any molecule.

Sample Problem 1 Calculate the molar mass of H 2O.

Solution H2O is the formula for water. It represents the fact that a water molecule contains 2 hydrogen atoms and 1oxygen atom. We make a chart to add up the mass of one mole of water by rewriting the formula vertically as follows:

relative number element atomic masses of atoms total

H 1.0 x 2 = 2.0O 16.0 x 1 = 16.0

1 mole H 2O = 18.0 grams H 2O

The molar mass of H 2O is 18.0 g/mole.

1 mole of oxygen atoms = 6.02 x 10 oxygen atoms

1 mole Ca = 40.1 grams Ca


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Sample Problem 2 Calculate the molar mass of Ca(NO 3)2.

SolutionCa 40.1 x 1 = 40.1

N 14.0 x 2 = 28.0O 16.0 x 6 = 96.0

1 mole Ca(NO 3)2 = 164.1 grams Ca(NO 3)2

The molar mass of Ca(NO 3)2 is 164.1 g/mole.

Using Moles in Calculations

You will remember from chapter 2 how we set up problems using dimensional analysis (factor label).

In this chapter we have learned three new equivalencies. Each can be expressed in fraction form to solvechemistry problems via dimensional analysis.

Sample Problem 3 What is the mass of 0.827 moles of H 2O?

Solution First find the molar mass of H 2O. Then use this as a conversion factor to get you from moles tograms.

0.827 molesH 2 O1

x18.0 gramsH 2 O

1 moleH 2 O= 14.9 gramsH 2 O

Sample Problem 4 How many moles of sodium hydroxide, NaOH, are there in 450. g of the compound?

Solution First find the molar mass of NaOH. Then use this as a conversion factor to get you from grams tomoles.

450. gramsNaOH 1

x1moleNaOH

40.0 gramsNaOH = 11.2 molesNaOH

Sample Problem 5 How many atoms are contained in 1.45 grams of Ca?

1.45 gramsCa1

X 1 moleCa

40.1 gramsCa X

6.02 x1023

atoms1 moleCa

= 2.18 x10 22 atomsCa

We are now ready to apply dimensional analysis/factor labelto mole calculations.

1 mole = 6.02 x 10 particles = 22.4 L of a gas at STP = molar mass of the element/compound


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Sample Problem 6 How many liters are contained in 3.29x10 24 molecules of carbon dioxide at STP?

Solution There is no need to calculate the molar mass of CO 2 for this problem. Molar masses are onlynecessary when starting with or ending with a mass.

Sample Problem 7 What is the mass of 45.0 L of carbon dioxide at STP?

Solution Find the molar mass of CO 2. Then use this as a conversion factor to get you between grams andmoles.

35.0 LCO21

x1moleCO 222.4 LCO2

x44.0 gramsCO 2

1moleCO 2= 68.8 gramsCO 2

Problems

A. Calculate the molar mass of each of the following:

1. KOH

2. CoCl 3

3. Ca(NO 3)2

4. Iron (III) oxide

5. Carbon tetrachloride

6. Sodium phosphate

7. Copper (II) nitrate

B. Calculate the number of moles in each of the following:

1. 35.7 L of N 2 at STP

2. 264 g of sodium hydroxide

3. 8.16x10 23 molecules of carbon tetrachloride

C. Calculate the mass of each of the following:

1. 2.70 moles of potassium chloride

2. 9.23x10 24 formula units of sodium chromate

3. 29.8 L of dinitrogen monoxide at STP

D. Calculate the number of atoms/molecules/formula units present in each of the following:

1. 26.0 g potassium nitrate

2. 0.83 moles iron

3. 16.9 L of sulfur dioxide at STP

3.29 x10 24 moleculesCO2

1 x

1 moleCO2

6.02 x10 23 moleculesCO2

x22.4 LCO

2

1 moleCO2

= 122 LCO2


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E. Calculate the number of liters present in each of the following.

1. 47.6 grams of sulfur hexafluoride at STP 3. 1.94 moles of carbon dioxide at STP

2. 5.10x10 18 atoms of argon at STP

F. Mixed Problems

1. You have 6.82 L of N 2 gas at STP. How many molecules is this?

2. You have 8.3 X 10 8 atoms of zinc. How many moles is this?

3. What is the volume of 52.0 grams of NH 3 gas at STP?

4. How many atoms would be found in 73,000 grams of Silver?

5. What is the volume of 0.847 grams of carbon dioxide at STP?

6. You have 618.5 grams of silver nitrate. How many moles is this?

7. What is the mass of 4.88 L of O 2 gas at STP?

8. What is the volume of 3.9 moles of N 2 gas at STP?

9. What is the mass of 1.74 X 10 16 formula units of sodium hydroxide?

10. What is the mass of 4.3 X 10 28 formula units of copper (II) nitrate?

11. You have 263 liters of dinitrogen pentoxide gas at STP. What is the mass of this gas?

12. What is the mass of 47,000,000 atoms of lithium?

13. What is the mass of 3,720 molecules of O 2 gas?

14. How many molecules are contained in 26 grams of water?15. How many formula units are contained in 0.285 grams of potassium hydroxide?

16. What is the volume of 4,260 atoms of Neon gas at STP?

17. What is the volume of 56.27 grams of nitrogen dioxide gas at STP?

18. What is the mass of 1 atom of iron?


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Answers to Problems

Part A1. 56.1 g/mole2. 165.4 g/mole3. 164.1 g/mole4. 159.6 g/mole5. 154.0 g/mole6. 164.0 g/mole7. 187.5 g/mole

Part B1. 1.59 moles N 2 2. 6.60 moles NaOH3. 1.36 moles CCl 4

Part C1. 201 g KCl2. 2480 g Na 2CrO 4 3. 58.5 g N 2O

Part D1. 1.55x10 23 formula units KNO 3 2. 5.0x10 23 atoms Fe3. 4.54x10 23 molecules SO 2

Part E1. 7.30 L SF 6 2. 1.89x10 -4 L Ar 3. 43.5 L CO 2

Part F1. 1.83x10 23 molecules N 2 2. 1.4x10 -15 moles Zn3. 68.5 L NH 3 4. 4.1x10 26 atoms Ag5. 0.431 L CO 2 6. 3.640 moles AgNO 3 7. 6.97 grams O 2 8. 87 L N 2 9. 1.16x10 -6 grams NaOH10. 1.3x10 7 grams Cu(NO 3)2 11. 1270 grams N 2O5 12. 5.4x10 -16 grams Li13. 1.98x10 -19 grams O 2 14. 8.7x10 23 molecules H 2O15. 3.06x10 21 formula units KOH16. 1.59x10 -19 L Ne17. 27.4 L NO 2 18. 9x10 -23 grams Fe

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