12.1 Test Review

What is percent composition?

• the percent (by mass) of all the elements in a compound

What is Avagadros’ constant – 6.02 X 1023

• the number of particles contained in 1 mole of a substance

What is the SI unit of chemical quantity?

• the mole

What is molar mass?

• the mass of 1 mole of a substance

classify the following substance as:

a. formula unit b. atomc. molecule

d. ion•Mg

•atom

classify the following substance as:

a. formula unit b. atomc. molecule

d. ion•C2O4

-2

•ion

classify the following substance as:

a. formula unit b. atomc. molecule

d. ion• Na(NO3)2

•formula unit

classify the following substance as:

a. formula unit b. atomc. molecule

d. ion• C6H12O6

•molecule

determine the molar mass for the following substance:

• Br

•80

determine the molar mass for the following substance:

• C6H12O6

•180

determine the molar mass for the following substance:

• Na(NO3)2

•147

Simplified mole conversion

How many moles are in 32 grams of SCl2?

32g SCl2 X 1

1 mole SCl2102 g SCl2

= 0.31 mol SCl2

Calculate the number of atoms in 3.2 moles of gold (Au)

3.2g mol Au X

1

6.02 X 10 23 atoms Au 1 mol Au

= 1.9 X 1024 atoms Au

What is the volume in liters of 12.4 mol of CO2 gas?

12.4 mol CO2 X

1

22.4 L CO2

1 mol CO2

= 277.8 L CO2

What is the volume in liters of 21.3g of N2 gas?

21.3 g N X 1

X 22.4 L N2

1 mol N2

= 17.04 L N2

1 mol N2

28 g N2

How many molecules are in 72.3 L of O2?

72.3 L O2 X 1

X 6.02 X 1023 O2

1 mol O2

= 1.9 X1024 molecules O2

1 mol O2

22.4 L O2

What is the mass in grams of 2.1 X 1021 formula units of MgCl2?

X 94 g MgCl2

1 mol MgCl2

2.1 X 1021 fu MgCl2 1

= 0.33 g MgCl2

X 1 mol MgCl26.02 X 1023 fu MgCl2

Determine the percent composition for the following compound: CaCl2

• Ca = 40.078 g• Cl2 = 2 x 35.453 g = + 70.906 g • CaCl2 = 110.984 g

Ca = 40.1g X 100 = 36.1% Ca 110.9Cl = 70.9g X 100 = 63.9% Cl 110.9g

empirical formula of a compound with percent composition of 38.4 % Mn, 16.8 % C, 44.7 % O

38.4 g Mn X 1mol Mn = 0.699 mole Mn 1 55 g Mn

16.8 g C X 1mol C = 1.339 mole C 1 12 g C

45.7 g O X 1mol O = 2.798 mole O 1 16 g O

0.699 mole Mn = 1 mol Mn0.699

1.339 mole C = 2 mol C 0.699

2.798 mole O = 4 mol O 0.699

So formula is MnC2O4

empirical and molecular formulas of a compoundwith percent composition of 71.0 % Cl, 25.0% C, 4.0 % H and mass 98.96 g/mol

71.0 g Cl X 1mol Cl = 2.0285 mole Cl 1 35 g Cl

25.0 g C X 1mol C = 2.0833 mole C 1 12 g C

4.0 g H X 1mol H = 4.0 mole H 1 1 g H

2.0285 mole Cl = 1 mol Cl 2.0285

2.0833 mole C = 1.02 mol C 2.0285

4.0 mole H = 1.97 mol H 2.0285

So formula is ClCH2

molecular mass of compound 98.96g = 2.01

molar mass of ClCH2 49 g

2 X (ClCH2) = Cl2C2H4