Malonic acid is used in the manufacture of barbiturates (sleeping pills). The composition of the acid is 34.6% C, 3.9% H, and 61.5% O. What is malonic acid’s empirical formula?

#### Solution:

### Assume a sample size of 100 g to make the math easier. Thus, we will assume there is a 100 g sample of malonic acid. Now, we will calculate the amount of grams that each element takes up in the sample size.

### Grams of carbon: 100 g \times 34.6% = 34.6 g

### Grams of hydrogen: 100 g \times 3.9% = 3.9 g

### Grams of oxygen: 100 g \times 61.5% = 61.5 g

### Now, we will convert each of these masses to moles by dividing by it’s specific molar mass.

### \text{Mol C} = 34.6\,g\,\text{C}\times \frac{1\text{mol C}}{12.01\, g\,\text{C}}

### \text{Mol C} =2.881\text{mol}

### \text{Mol H} = 3.9\,g\,\text{H}\times \frac{1\text{mol H}}{1.008\, g\,\text{H}}

### \text{Mol H} =3.87\text{mol}

### \text{Mol O} = 61.5\,g\,\text{O}\times \frac{1\text{mol O}}{15.999\, g\,\text{O}}

### \text{Mol O} =3.844\text{mol}

### To figure out the integers of the empirical formula, simply divide each value by the smallest mol number we found, which was 2.881.

### C= \frac{2.881}{2.881}=1

### H=\frac{3.87}{2.881}=1.34\approx \frac{4}{3}

### O=\frac{3.884}{2.881}= 1.334\approx \frac{4}{3}

### Notice we got fractional values. To get whole numbers, we will multiply these values by the lowest common denominator. In this case, the lowest common denominator is 3.

### C= 1\times 3=3

### H=\frac{4}{3}\times 3=4

### O=\frac{4}{3}\times 3=4

### Thus, the empirical formula is C_3H_4O_4