Question

A compound is 21.6% Mg, 21.4% C, and 57.0% O. What is the empirical formula of the compound? ​

Answer 1

Answer:

Compound x = $MgC_{2}O_{4}$

Explanation:

Let the compound be x

Assuming we have a 100g of compound x

Given the following data;

Magnesium, Mg = 21.6% = 21.6g

Carbon, C = 21.4% = 21.4g

Oxygen, O = 57.0% = 57.0g

Now, we would find the amount of moles for each element.

Atomic mass of Mg = 24.30g

Atomic mass of C = 12.01g

Atomic mass of O = 16.00g

Amount of moles for Mg;

21.6*(1/24.30) = 0.89mol

Amount of moles for C;

21.4*(1/12.01) = 1.78mol

Amount of moles for O;

57.0*(1/16.00) = 3.56mol

We then divide by the smallest to find the ratio;

0.89/0.89 = 1 Mol of Mg

1.78/0.89 = 2 Mol of C

3.56/0.89 = 4 Mol of O

Therefore, the ratio of Mg, C and O is 1:2:4.

Compound x = $MgC_{2}O_{4}$

Hence, the empirical formula of the compound is $MgC_{2}O_{4}$