Question

What is the empirical formula of a compound that contains 43.38% sodium, 11.33% carbon, and 45.29% oxygen?

Answer 1

Answer: $Na_2CO_3$.

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 Na = 43.38 g

Mass of C = 11.33 g

Mass of O = 45.29 g

Step 1 : convert given masses into moles.

Moles of Na=$\frac{\text{ given mass of Na}}{\text{ molar mass of Na}} \frac{43.38g}{23g/mole}=1.89moles$

Moles of C =$\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{11.33g}{12g/mole}=0.94moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{45.29g}{16g/mole}=2.83moles$

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

For Na = $\frac{1.89}{0.94}=2$

For C =$\frac{0.94}{0.94}=1$

For O =$\frac{2.83}{0.94}=3$

The ratio of Na : C: O = 2: 1: 3

Hence the empirical formula is $Na_2CO_3$.

Answer 2

What is the empirical formula of a compound that contains 43.38% sodium, 11.33% carbon, and 45.29% oxygen?


Na2CO3