Answer:
[tex]\boxed{\text{CH$_{2}$}}[/tex]
Explanation:
1. Calculate the mass of each element
[tex]\text{Mass of C} = \text{2.67 g } \text{CO}_{2}\times \dfrac{\text{12.01 g C}}{\text{44.01 g }\text{CO}_{2}}= \text{0.7286 g C}\\\\\text{Mass of H} = \text{1.10 g }\text{H$_{2}$O}\times \dfrac{\text{2.016 g H}}{\text{18.02 g } \text{{H$_{2}$O}}} = \text{0.1231 g H}[/tex]
2. Calculate the moles of each element
[tex]\text{Moles of C = 0.7286 g C}\times\dfrac{\text{1 mol C}}{\text{12.01 g C }} = \text{0.060 67 mol C}\\\\\text{Moles of H = 0.1231 g H} \times \dfrac{\text{1 mol H}}{\text{1.008 g H}} = \text{0.1221 mol H}[/tex]
3. Calculate the molar ratios
Divide all moles by the smallest number of moles.
[tex]\text{C: } \dfrac{0.06067}{0.06067}= 1\\\\\text{H: } \dfrac{0.1221}{0.06067} = 2.015[/tex]
4. Round the ratios to the nearest integer
C:H = 1:2
5. Write the empirical formula
The empirical formula is [tex]\boxed{\text{CH$_{2}$}}[/tex]
