Question

27) a) Find the empirical and molecular formula of the compound containing
se k 51.28% carbon, 9.40% Hydrogen, 27.35% Oxygen and 11.97% Nitrogen.
Molar mass of the compound is 234 gm/mol.
(Atomic mass of C = 12. H= 1 0.= 16 N = 14)​

Answer 1

Answer:

One strategy to use when given the molar mass of the compound is to pick a sample that would correspond to one mole and use it to find the molecular formula without finding the empirical formula first.

In this case, a molar mass of

88 g mol

−

1

tells you that one mole of this compound has a mass of

88 g

.

Use the percent composition of the compound to determine how many grams of each element you'd get in this

88-g

sample. Remember, you can convert between percentages and grams by using a sample of

100 g

.

You will thus have

For C:

88

g sample

⋅

54.53 g C

100

g sample

=

47.986 g C

For H:

88

g sample

⋅

9.15 g H

100

g sample

=

8.052 g H

For O:

88

g sample

⋅

36.32 g O

100

g sample

=

31.962 g O

Now use the molar mass of each element to determine how many moles of each are present in one mole of this compound

For C:

47.986

g

⋅

1 mole C

12.011

g

=

3.995

≈

4 moles C

For H:

8.052

g

⋅

1 mole H

1.00794

g

=

7.990

≈

8 moles H

For O:

31.962

g

⋅

1 mole O

15.9994

g

=

1.998

≈

2 moles O

Since this is how many moles of each element are present in one mole of this compound, you can say that one molecule of the compound will contain

4

atoms of carbon

8

atoms of hydrogen

2

atoms of oxygen

Therefore, the compound's molecular formula, which tells you exactly how many atoms of each element make up one molecule of the compound, will be

C

4

H

8

O

2