Empirical Formula Chemistry Example 4

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Example 4

medium
A compound is 40.0%40.0\% carbon, 6.7%6.7\% hydrogen, and 53.3%53.3\% oxygen by mass. What is its empirical formula?

Solution

  1. 1
    Assume 100100 g of compound: C = 40.040.0 g, H = 6.76.7 g, O = 53.353.3 g. Convert to moles: C =40.012.01≈3.33= \frac{40.0}{12.01} \approx 3.33, H =6.71.008≈6.65= \frac{6.7}{1.008} \approx 6.65, O =53.316.00≈3.33= \frac{53.3}{16.00} \approx 3.33.
  2. 2
    Divide each by the smallest value, 3.333.33: C =1= 1, H =2= 2, O =1= 1. The empirical formula is CH2O\text{CH}_2\text{O}.

Answer

CH2O\text{CH}_2\text{O}
Empirical formulas use the lowest whole-number mole ratio. Percent composition problems are solved by assuming a 100 g sample, converting to moles, and simplifying the ratio.

About Empirical Formula

The chemical formula that shows the simplest whole-number ratio of atoms of each element present in a compound, obtained by dividing all subscripts by their.

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More Empirical Formula Examples

Example 1 medium

A compound contains [formula] C, [formula] H, and [formula] O by mass. Determine the empirical formu

Example 2 hard

A compound has empirical formula [formula] and a molar mass of [formula]. Find the molecular formula

Example 3 medium

A compound is [formula] Mg and [formula] N. Find the empirical formula.

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