Stoichiometry

Part 3!

Mass to Mole Conversions

Remember you CANNOT convert directly from mass!

If you begin with mass, you must convert to moles FIRST!

Then compare moles to moles using coefficients.

Mass → conversion to moles → mole ratio

Sample Problem 9-4

The catalytic oxidation of ammonia is run using 824 g of ammonia and excess oxygen.

How many moles of NO are formed? First, write the balanced equation:

NH3 + O

2 → NO + H

2O

4NH3 + 5O

2 → 4NO + 6H

2O

4NH3 + 5O

2 → 4NO + 6H

2O

Convert from 824 g NH3 to moles:

824 g NH3 x 1 mole NH3 = 48.3839525 mole

17.03044 g Use the mole ratio to find moles of NO:

48.3839525 mole NH3 x 4 mole NO = 48.3839525

4 mole NH3

Check significant digits! 48.4 mole NO

4NH3 + 5O

2 → 4NO + 6H

2O

How many moles of H2O are formed from 824 g

of NH3?

824 g NH3 = 48.4 mole NH

3

Then use mole ratio: 48.4 mole NH3 x 6 mole H

2O = 72.57592875 mole

4 mole NH3

Correct for significant digits: 72.6 mole H2O

Mass-mass calculations

First convert from mass to moles Then use mole ratio Then convert from moles to mass Remember, you CANNOT convert directly from

mass of one substance to mass of another!

Sample problem 9-5

The reaction of tin with hydrogen fluoride produces tin (II) fluoride and hydrogen gas

How many grams of SnF2 are produced from

the reaction of 30.00 g of HF with Sn? First, write the balanced equation:

Sn + HF → SnF2 + H

2

Sn + 2HF → SnF2 + H

2

Sn + 2HF → SnF2 + H

2

30.00 g HF x 1 mole HF = 1.499557 mole HF 20.006 g HF

1.499557 mole HF x 1 mole SnF2 = 0.74978

2 mole HF mole SnF2

0.74978 mole SnF2 x 156.706 g SnF

2 =117.49

1 mole SnF2 g SnF

2

Correct for significant digits: 117.5 g SnF2

Now it's your turn!

Do practice problems 1-2 page 285 and practice problems 1-3 page 287.