Question

An unknown compound contains only the three elements C, H, and O. A pure sample of the compound is analyzed and found to be 65.60 percent C and 9.44 percent H by mass. (a) Determine the empirical formula of the compound.

Answer 1

Answer: The empirical formula is $C_7H_{12}O_2$

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= 65.60 g

Mas of H = 9.44

Mass of O = (100-(65.6+9.44)= 24.96 g

Step 1 : convert given masses into moles.

Mass of C =$\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{65.5g}{12g/mole}=5.47moles$

Moles of H =$\frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{9.44g}{1g/mole}=9.44moles$

Moles of O=$\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{24.96g}{16g/mole}=1.56moles$

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

For C = $\frac{5.47}{1.56}=3.5$

For H = $\frac{9.44}{1.56}=6$

For O =$\frac{1.56}{1.56}=1$

The ratio of C:H:O= 3.5: 6: 1

Converting them into whole numbers, the ratio will be 7: 12: 2

Hence the empirical formula is $C_7H_{12}O_2$