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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?

ACHO
BC4H6O10
CC2HNO5
DCannot determine from the information provided

Verified Answer :

MissPhiladelphia

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

Answer: C. \(C_2HNO_5\)

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

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

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

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

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

For C = \(\frac{1.7}{0.8}=2\)

For H =\(\frac{1}{0.8}=1\)

For N =\(\frac{0.8}{0.8}=1\)

For O =\(\frac{4.2}{0.8}=5\)

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

Hence the empirical formula is \(C_2HNO_5\)

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