1. a. sulfur in 3.54 g H2S % comp. S is: 32.1 x 100 = 94.1% 34.1
#64 Using your answers from Problem 63, calculate the
number of grams of these elements.
a. sulfur in 3.54 g H2S
% comp. S is: x 100 = 94.1%
34.1
Therefore the number of grams of sulfur is:
.941 x 3.54 = g Sulfur
B. nitrogen in 25.0 g (NH4)2C2O4
% comp. N is: x 100 = 22.6%
124.0
Therefore the number of grams of nitrogen is:
.226 x 25.0 = g N

2. % comp. P is: 31.0 x 100 = 18.9% 164.0 c. magnesium in 97.4 g Mg(OH)2
#64 Using your answers from Problem 63, calculate the
number of grams of these elements.
c. magnesium in 97.4 g Mg(OH)2
% comp. Mg is: x 100 = 41.7%
82.6
Therefore the number of grams of magnesium is:
.417 x 97.4 = g Magnesium
d. phosphorus in 804 g Na3PO4
% comp. P is: x 100 = 18.9%
164.0
Therefore the number of grams of nitrogen is:
.189 x = 152 g P

3. #65 Which of the following compounds has the highest iron content?
a. FeCl2
% composition Fe is: x 100 = 44.0%
126.8
% composition Fe is: x 100 = 32.3%
55.8+(12x6)+9+96
b. Fe(C2H3O2)3
% composition Fe is: x 100 = 62.1%
89.8
c. Fe(OH)2
% composition Fe is: x 100 = 77.7%
71.8
d. FeO

4. continued next slide #66 You find that 7.36 g of a compound
has decomposed to give 6.93 g of oxygen.
The only other element in the compound
is hydrogen. If the molar mass of the compound
is 34.0 g/mole, what is its molecular formula?
1. Go from grams to an % composition first:
6.93 x 100 = 94.2 % Oxygen
7.36
x 100 = 5.8 % Hydrogen
7.36
continued next slide

5. continued next slide 94.2 % Oxygen 5.8 % Hydrogen
2. Assume a 100g sample gives
94.2 g O
5.8 g H
3. Convert to moles:
x 1 mole = 5.88 moles
16.0 g O
x 1 mole = 5.88 moles
1.0 g H
continued next slide

6. 4. divide by the smaller of the two
numbers: (in this case the numbers are the
same:
5.88 moles O/5.88 = 1
5.88 moles H/ 5.88 = 1
The empirical formula (lowest whole number ratio)
is HO or a 1 to 1 ratio
5. Since the problem states the molar mass is
34.0 you have to multiply the ratios
until you get 34.1 grams for a molecular formula:
continued next slide

7. Therefore the molecular formula is H2O2
HO is g/mole
H2O2 is g/mole which matches our problem
Therefore the molecular formula
is H2O2

8. NO because you can reduce this ratio to S1Cl1
#67 Which of the following can be classified
as an empirical formula?
a. S2Cl2
NO because you can reduce
this ratio to S1Cl1
Therefore it is not the smallest whole number
ratio.
b. C6H10O4
NO because you can reduce
this ratio to C3H5O2
Therefore it is not the smallest whole number
ratio.
continued next slide

9. reduce down any further Ionic compounds are already
#67 Which of the following can be classified
as an empirical formula?
c. Na2SO4
YES because you cannot
reduce down any further
Ionic compounds are already
given in their empirical
formula.

10. to get this you just multiply the molar
#68. What is the molecular formula for each
compound? Each compound’s empirical
formula and molar mass is given.
a. CH2O 90 g/mole
to get this you just multiply the molar
mass of the given formula by small
whole numbers until you get the answer:
CH2O is 30g/mole. Therefore this x 3
gives the 90 g/ mole. The molecular
formula is:
C3H6O3

11. this is tricky because it appears to be
#68. What is the molecular formula for each
compound? Each compound’s empirical
formula and molar mass is given.
b. HgCl g/mole
this is tricky because it appears to be
an ionic compound but it is molecular
therefore the molar mass of the given
empirical formula is: 236.1g/mole
multiplying by 2 given therefore
the molecular formula is
Hg2Cl2

12. #69 Determine the molecular formula
for each compound.
a % O
5.9 % H
molar mass: 34g
1. Assume a 100 g sample:
94.1g O
5.9g H
continued next slide

13. So we see a ratio of 1 to 1 thus the empirical formula is HO
2. convert to moles:
94.1g O x 1 mole = 5.88
16.0 g O
5.9g H x 1 mole = 5.9
1.0 g H
3. divide by the smaller number to get
a ratio:
5.88/5.88 = 1
5.9/ = 1
So we see a ratio
of 1 to 1 thus the
empirical formula
is HO
continued next slide

14. The molecular formula is a multiple
of HO which has a molar mass of
17.0
The original problem stated that the
molecule has a molar mass of 34 g
Therefore we can see that the empir-
ical formula times 2 gives:
17.0 x g The formula is:
H2O2

15. #69b Determine the molecular formula
for each compound.
b % C
4.2 % H
45.1 % O
molar mass: 142g
The book is wrong
1. Assume a 100 g sample:
50.7g C
4.2g H
45.1 g O
continued next slide

16. 45.1g O x 1 mole = 2.84 16.0 g H 2. convert to moles:
50.7g C x 1 mole = 4.22
12.0 g O
4.2g H x 1 mole = 4.2
1.0 g H
45.1g O x 1 mole =
16.0 g H
continued next slide

17. x 2 = 3 2.84/2.84 = 1 x 2 = 2 3. divide by the smaller number to get
a ratio:
4.22/2.84 = 1.48
4.2/ =
2.84/ = 1
x 2 = 3
x 2 = 2
These numbers are almost right,
but they need to be whole numbers so multipy all
by 2. Rounding you get C3H3O2
continued next slide

18. C6H6O4 which has a molar mass of:
4. Finally, taking the empirical
formula and multiplying the subscipts
until I get the molar mass of the
original problem (142g/mole) you get:
C3H3O2 is molar mass of: 71g/mole
times 2 is
C6H6O4 which has a molar mass of:
142 which is the answer!