We are asked to determine the molecular formula of a compound with a molar mass of 180.18 g/mol and a percentage composition of 39.99% C, 6.73% H, 53.28% O.

**This means we need to do the following steps:**

*Step 1:* Calculate the mass and moles of C, H, and O in the compound.

*Step 2:* Determine the lowest whole number ratio of C, H, and O to get the empirical formula.

*Step 3:* Get the ratio of the molar mass and empirical mass to determine the molecular formula.

**Step 1:** The compound is composed of C, H, O and we’re given the mass percent of **C (39.99% C)**, **H (6.73% H) and O (53.28% O)**. Check if the mass percentages of a compound add up to 100%.

This means:

$\overline{){\mathbf{Total}}{\mathbf{}}{\mathbf{mass}}{\mathbf{\%}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}{\mathbf{mass}}{\mathbf{\%}}{\mathbf{}}{\mathbf{C}}{\mathbf{}}{\mathbf{+}}{\mathbf{}}{\mathbf{mass}}{\mathbf{\%}}{\mathbf{H}}{\mathbf{}}{\mathbf{+}}{\mathbf{}}{\mathbf{mass}}{\mathbf{\%}}{\mathbf{}}{\mathbf{O}}}\phantom{\rule{0ex}{0ex}}\mathbf{Total}\mathbf{}\mathbf{mass}\mathbf{}\mathbf{\%}\mathbf{}\mathbf{=}\mathbf{}\mathbf{39}\mathbf{.}\mathbf{99}\mathbf{\%}\mathbf{}\mathbf{+}\mathbf{}\mathbf{6}\mathbf{.}\mathbf{73}\mathbf{\%}\mathbf{+}\mathbf{}\mathbf{53}\mathbf{.}\mathbf{28}\mathbf{\%}\phantom{\rule{0ex}{0ex}}\mathbf{Total}\mathbf{}\mathbf{mass}\mathbf{\%}\mathbf{}\mathbf{=}\mathbf{}\mathbf{100}\mathbf{.}\mathbf{00}\mathbf{\%}$

Recall that * mass percent *is given by:

$\overline{){\mathbf{\%}}{\mathbf{mass}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}\frac{\mathbf{mass}\mathbf{}\mathbf{X}}{\mathbf{Total}\mathbf{}\mathbf{mass}}{\mathbf{x}}{\mathbf{100}}}$

Assuming we have **100 g** of the compound, this means we have ** 39.99 g C, 6.73 g H, and 53.28 g O**.

Now, we need to get the moles of each element in the compound.

The atomic masses are **12.01 g/mol C, 1.00 g/mol H, 15.99 g/mol O.**

A compound has a molar mass of 180.18 g/mol. Given the following percent composition, calculate the molecular formula.

39.99% C, 6.73% H, 53.28% O

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