# Chapter 7

Molecular mass is the sum of the atomic masses of the atoms of each element in a molecule.

Mole - the measure of quantity of a substance.

1 mole of any substance has the amu of that substance

1 mole of any substance has 6.0225 X 1023 atoms, molecules, ions of that substance

amu of a substance = 1 mole of that substance = 6.0225 X 1023

Change moles to grams:

Change moles to grams:

 0.62 mol NaOH 40.01 g NaOH 1 mol NaOH

Change grams to moles:

 2.5 g NaOH 1 mol NaOH 40.01 g NaOH

Change moles to atoms/molecules:

 0.62 mol NaOH 6.02 x 1023 atoms/mlc NaOH 1 mol NaOH

Change atoms/molecules to moles:

 2.35 X 1022 atoms/mlc NaOH 1 mol NaOH 6.02 X 1023 atoms/mlc NaOH

Change from atoms/molecules to grams:

 2.35 X 1022 atoms/mlc NaOH 1 mol NaOH 40.01 g NaOH 6.02 X 1023 atoms/mlc NaOH 1 mol NaOH

Change from grams to atoms/molecules:

 2.35 g NaOH 1 mol NaOH 6.02 x 1023 atoms/mlc NaOH 40.01 g NaOH 1 mol NaOH

Percent Composition

Calculate the percent composition of Ba(OH)2.

1. Calculate the amu of the compound.
```
Ba  =  1   x  137.33  =  137.33
O  =  2   x   16.0   =   48.0
H  =  2   x    1.01  =    2.02
_________
187.35```
2. Divide the total amu of each element by the total amu for the compound.

```
137.33 ÷ 187.35  =  0.733 x 100 = 73.3% Ba

48.0  ÷  187.35  =  0.256  x  100  =  25.6% O

2.02  ÷  187.35  =  0.011  x  100  =  1.1% H```

Empirical and Molecular Formulas

```
Change % to grams               Change grams to moles               Divide by the lowest               Multiply to get whole #'s

63.1% Mn       =  63.1 g Mn  x 1 mol Mn     =     1.15 ÷ 1.15  =  1          (not necessary in this example)
54.94 g Mn

36.9% S         =  36.9 g S   x   1 mol S      =      1.15 ÷ 1.15  =  1          (not necessary in this example)
32.07 g S

Empirical Formula  =  MnS```
Use the same procedures for the molecular formula and add the last steps.
```
Change % to grams               Change grams to moles               Divide by the lowest               Multiply to get whole #'s

92.25% C       =  92.25 g C  x 1 mol C     =     7.68 ÷ 7.67  =  1          (not necessary in this example)
12.01 g C

7.75% H         =  7.75 g H   x   1 mol H     =    7.67 ÷ 7.67  =  1          (not necessary in this example)
1.01 g H

Empirical Formula  =  CH

molecular mass  =  78

[EFamu]x   =   molecular mass

[13.02]x   =   78    (solve for x)

13.02x  =  78
x  =  6

plug in x and multiply through the empirical formula subscripts

[CH]6      =    C6H6  =  molecular formula```

Practice Problems:

A chemist has a jar containing 388.2 g of iron filings. How many moles of iron does the jar contain?

A student needs 0.366 mol of zinc for a reaction. What mass of zinc in grams should the student obtain?

How many moles of lithium are there in 1.204 x 1024 lithium atoms?

How many boron atoms are there in 2.00 g of boron? atoms B)

How many moles of carbon dioxide are in 66.0 g of dry ice, which is solid CO2?

Determine the number of molecules in 0.0500 mol of hexane, C6H14.

Calculate the mass of 8.25 x 1022 mlc of bromine pentafluoride.

Calcule the number of atoms in 22.9 g of sodium sulfide.

Calculate the number of grams in 3.0 x 1021 mlc of CH2CHCN.