# Using Molar Masses 24.31 g 0.475 mol x = 11.5 g Mg

```Using Molar Masses
A reaction requires 0.475 mol of Magnesium turnings. How much
magnesium, in g, is that?
24.31 g
0.475 mol x
= 11.5 g Mg
1 mol
“1” always goes with moles, and the number
off of the periodic table goes with grams
How many atoms of sodium are in 13.0 g of sodium metal?
1 mole
6.022 x 10 23 atoms
13.0 g Na x
x
= 3.41 x 10 23 atoms Na
22.998 g
1 mole
For an experiment that I am performing, I need to add 12.5 mmol
of iron (II) sulfate [FeSO4]. How many grams must I weigh out?
1 mol
151.92 g
12.5 mmol x
x
= 1.90 g
1000 mmol
1 mole
Practice Problems
What is the mass of 4.91 x 1021 platinum atoms?
4.91 x 10 21 atoms Pt x
1 mole
195.08 g
x
= 1.59 g Pt
23
1 mole
6.022 x 10 atoms
How many atoms of copper are in a 133 kg pure copper statue?
1000 g
1 mol
6.022 x 10 23 atoms Cu
133 kg Cu x
x
x
= 1.26 x 10 27 atoms Cu
1 kg
63.546 g
1 mole
Determine the number of sulfur atoms in 1.59 mmol of carbon
disulfide.
1 mol
6.022 x 10 23 molecules
2 atoms S
1.59 mmol CS2 x
x
x
= 1.91 x 10 21
1000 mmol
1 mol
1 molec CS2
Determine the number of H atoms are in 9.88 mol of NH3.
6.022 x 10 23 molecules
3 atoms H
9.88 mol NH3 x
x
= 1.78 x 10 25 atoms H
1 mole
1 molec NH3
Mole-to-Mole Comparisons
A reaction requires 10.0 g of iron atoms. How many grams of iron
(II) hydroxide are required to complete the reaction?
1 mol Fe(OH)2
1 mol
89.866 g
10.0 g Fe atom x
x
x
= 16.1 g Fe(OH)2
55.85 g
1 mol Fe
1 mol
Mass Percents: Determine the mass percent of oxygen in FeSO4. Use
that value to determine the grams of oxygen in 98.75 g of FeSO4.
Mass percent =
4 x mass Oxygen
4 x 15.9996 amu
x 100 =
x 100 = 42.126% O
mass FeSO 4
151.92 amu
98.75 g FeSO 4 x 0.42126 = 41.60 g O
Empirical Formula
The pre-hormone androstenedione (commonly called andro) has been in
the sports news in recent years owing to its arguable contribution to the
breaking of the homerun record. The product has a composition that is
79.68% carbon (C), 9.15% hydrogen (H), and 11.17% oxygen by weight.
What is the empirical formula for this compound?
assume 100 g: all % ! g
1 mol
79.68 g C x
= 6.63 mol
= 9.5 x 2 = 19
0.698 mol
12.011 g
1 mol
9.15 g H x
= 9.08 mol
= 13 x 2 = 26
0.698
mol
1.0079 g
1 mol
11.17 g O x
= 0.698 mol
=1x2=2
0.698 mol
15.9996 g
The empirical formula is C19 H 26 O 2
Empirical Formula
Determine the empirical formula for acetominophen, the active
ingredient in Tylenol®, from the elemental analysis.
C 63.56%, H 6.00%, N 9.27%, O 21.17%
assume 100 g: all % ! g
1 mol
63.56 g C x
= 5.29 mol
=8
0.662 mol
12.011 g
1 mol
6.00 g H x
= 5.95 mol
=9
0.662
mol
1.0079 g
1 mol
9.27 g N x
= 0.662 mol
=1
0.662 mol
14.0067 g
1 mol
= 1.32 mol
=2
0.662 mol
15.9996 g
The empirical formula is C8 H 9 NO 2
21.17 g O x
Empirical Formula
Determine the empirical formula for glucose.
C 40.00%, H 6.72%, O 53.29%
assume 100 g: all % ! g
1 mol
40.00 g C x
= 3.33 mol
=1
3.33
mol
12.011 g
1 mol
6.72 g H x
= 6.67 mol
=2
3.33 mol
1.0079 g
1 mol
53.29 g O x
= 3.33 mol
=1
3.33
mol
15.9996 g
The empirical formula is CH 2 O
Empirical Formula → Molecular Formula
Use the empirical formula and the molar mass (given) to find the
molecular formula for glucose.
The molecular formula for glucose is some multiple of the
empirical formula: (CH2O)n
molar mass
n=
formula weight (molar mass of the emp. formula)
180.2 amu
=6
30.0 amu
glucose = (CH 2 O)6 = C6 H12 O6
n=
Determining Molecular Formula
Adipic acid is used in the commercial manufacture of Nylon. The
composition of the acid is 49.3% C, 6.9% H, and 43.8% O by
mass. The molecular weight is 146 amu. What is the molecular
1 mol
= 4.10 mol
= 1.50 x 2 = 3
2.74
mol
12.011 g
1 mol
6.9 g H x
= 6.85 mol
= 2.50 x 2 = 5
2.74
mol
1.0079 g
1 mol
43.8 g O x
= 2.74 mol
=1x2=2
2.74
mol
15.9996 g
The empirical formula is C3 H 5O 2 ! The formula weight = 73.07 amu
49.3 g C x
146 amu
=2
73.07 amu
The molecular formula is (C3 H 5O 2 )2 = C6 H10 O 4
n=
```