Question

A compound is found to be 20 percent carbon, 1 percent hydrogen, and 12 percent nitrogen, with the rest being oxygen. What is the empirical formula of this compound?

a. CHO
b. C4H6O10
C. C2HNO5
d. Cannot determine from the information provided

Answer 1

Answer: C.
`C_2HNO_5`

Step-by-step explanation:

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C = 20 g

Mass of H= 1 g

Mass of N= 12 g

Mass of oxygen = 100-(20+1+12)= 67 g

Step 1 : convert given masses into moles.

Moles of C =
`\frac{\text{ given mass of C}}{\text{ molar mass of C}}= (20g)/(12g/mole)=1.7moles`

Moles of H =
`\frac{\text{ given mass of H}}{\text{ molar mass of H}}= (1g)/(1g/mole)=1mole`

Moles of N =
`\frac{\text{ given mass of N}}{\text{ molar mass of N}}= (12g)/(14g/mole)=0.8moles`

Moles of O =
`\frac{\text{ given mass of O}}{\text{ molar mass of O}}= (67g)/(16g/mole)=4.2moles`

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C =
`(1.7)/(0.8)=2`

For H =
`(1)/(0.8)=1`

For N =
`(0.8)/(0.8)=1`

For O =
`(4.2)/(0.8)=5`

The ratio of C : H : N : O= 2 : 1 : 1 : 5

Hence the empirical formula is
`C_2HNO_5`

Answer 2

First we need to determine the amount of oxygen present.

%O = 100 - 20 -1 -12 = 67%

Thus

C = 20%
H =1%
N= 12%
O = 67%

since we are not given if the given percentages are in mass or mole, we will assume that it is by mass. therefore with a basis of 100 g of compound the mole percentages are

C = 0.22
H = 0.13
N = 0.11
O = 0.54

dividing everything by the lowest percentage (0.11) the empirical formula is

C2HNO5