Chapter 6

Question 1

What is a mole?

Answer 1

It is a way to count atoms or molecules.

Question 2

How many atoms or molecules in a mole?

Answer 2

6.022 X 10²³

Question 3

What is Avogadro’s number?

Answer 3

6.022 X 10²³

Question 4

What is Avogadro’s number used to do?

Answer 4

convert between number ot atoms, molecules, or formula units.

Question 5

What is the molar mass of an element?

Answer 5

It is the mass of one mole of atoms of the element.

Question 6

What can the molar mass be used to do?

Answer 6

Used to convert the mass of an element to the number of moles.

Question 7

How is the mass of an element converted to the number of atoms.

Answer 7

Example: 24.02 g of carbon with 12.01 amu

24.02 g of C x ( 6.022 x 10²³ atoms of C / 12.01 g of C) = 2 moles or 2.0 x 6.022x 10²³ molecules

Question 8

What is molar mass equivalent to?

Answer 8

Formula mass in atomic mass units. For example for carbon: 12.01 amu = 12.01 grams/mole

Question 9

What is the molar mass of nitrogen dioxide (NO₂) if N = 14.01 amu and O = 16.00 amu?

Answer 9

1 x 14.01 amu + 2 x 16.00 amu = 46.01 amu = 46.01 grams or NO₂ per mole

Question 10

If there is 92.02 grams of NO₂, how many moles are there is N = 14.01 amu and O = 16.00 amu?

Answer 10

For NO₂, molecular mass is 1 x 14.01 amu + 2 x 16.00 amu = 46.01 amu = 46.01 grams/mole.

So (92.02 grams) / (46.01 grams/mole) = 2 moles

Question 11

If there is 6.005 g of carbon (C = 12.01 amu), how many molecules are there?

Answer 11

6. 005 grams / (12.01 grams/mole) = 0.5 moles
7. 5 moles * (6.022 x 10²³ molecules/mole) = 3.011 x 10²³ molecules

Question 12

What is the empirical formula?

Answer 12

Smallest whole number ratio of each type of atom in molcule.

Question 13

What is the molecular formula?

Answer 13

The actual number of each atom in a molecule.

Question 14

What is the molar mass?

Answer 14

It is a conversion factor between the mass (in grams) and the moles. For example, carbon is 12.01 grams/mole.

Question 15

What is the mass percent?

Answer 15

It is the element’s percent of the total mass of the compound.

Question 16

True or False: mass percent can be used to determine the empirical formula if the elements are known

Answer 16

True. With elements’ molar mass, number of moles can be determined. Once moles are known then empirical formula is just lowest whole number ratio.

Question 17

How many atoms are there in 0.5 moles?

Answer 17

0.5 x 6.022 x 10²³

Question 18

If carbon is 12.01 grams/mole, how many moles are there in 24.02 grams

Answer 18

24.02 grams / (12.01 grams/mole) = 2 moles

Question 19

If helium is 4.00 grams/mole, how much does 6.022 x 10²⁴ atoms weigh?

Answer 19

6.022 x 10²⁴ atoms / (6.022 x 10²³ atoms/mole) x 4.00 grams/mole = 400 grams

Question 20

What is mole mass of bicarbonate (HCO3) if H = 1.01 g/mole, C = 12.01 g/mole, O = 16.00 g/mole

Answer 20

1 x 1.01 g/mole + 1 x 12.01 g/mole + 3 x 16.00 g/mole = 61.02 g/mole

Question 21

What is the empirical formula of the compound which contains 2.02 g of hydrogen, 48.04 g of carbon, and 32.00 g of oxygen. H = 1.01 g/mole, C = 12.01 g/mole, O = 16.00 g/mole.

Answer 21

H: 2.02 g / 1.01 g/mole = 2 moles

C: 48.04 g / 12.01 g/mole = 4 moles

O: 32.00 g / 16.00 g/mole = 2moles

Formula: H₂C₄O₂

Empirical Formula: HC₂O

Question 22

What is mass percent of oxygen in acetic acid (C₂H₄O₂) if H = 1.01 g/mole, C = 12.01 g/mole, O = 16.00 g/mole

Answer 22

Molar mass of C₂H₄O₂ is 60.06 g/mole and O₂ is 32.00 g/mole. Mass percent of O₂ is 32.00/60.06 = 0.5328 = 53.28%