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3 years ago

A compound is found to have an empirical formula of CH2O. If its molecular mass is 60.0 g/mol, what is its molecular formula?

1 answer:

8 0

The empirical formula CH₂O has a mass [(12 × 1) + (1 × 2) + (16 × 1)] = 30 g/mol

If the empirical formula is 30 g/mol,
and the molecular formula is 60 g/mol

Then the multiple is = 60 g/mol ÷ 30 g/mol
= 2

Therefor the molecular formula is 2(CH₂O) = C₂H₄O₂ (OPTION 2)

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3 years ago

Read 2 more answers

quien y en que cantidad sera el reactivo limite, si utilizamos 125 g de ácido, H (CH3COO) y 275 g de hidróxido , Al(OH)3

Deffense [45]

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The limiting reactant is acetic acid. All 125 g will react.

Explanation:

1. Assemble the information

We will need a balanced equation with masses and molar masses, so let’s gather all the information in one place.

Mᵣ: 60.05 78.00

3CH₃COO-H + Al(OH)₃ ⟶ (CH₃COO)₃Al + 3H₂O

Mass/g: 125 275

2. Calculate the moles of each reactant

$\text{Moles of CH$_{3}$COOH} = \text{125 g CH$_{3}$COOH} \times \dfrac{\text{1 mol CH$_{3}$COOH}}{\text{60.05 g CH$_{3}$COOH}} = \text{2.082 mol CH$_{3}$COOH }\\\\\text{Moles of Al(OH)}_{3} = \text{275 g Al(OH)}_{3} \times \dfrac{\text{1 mol Al(OH)}_{3}}{\text{78.00 g Al(OH)}_{3}} = \text{3.526 mol Al(OH)}_{3}$

3. Calculate the moles of (CH₃COO)₃Al from each reactant

$\textbf{From CH$_{3}$COOH:}\\\text{Moles of (CH$_{3}$COO)$_{3}$Al} = \text{2.082 mol CH$_{3}$COOH} \times \dfrac{\text{1 mol (CH$_{3}$COO)$_{3}$Al}}{\text{3 mol CH$_{3}$COOH}}\\\\= \text{0.6939 mol (CH$_{3}$COO)$_{3}$)Al}\\\textbf{From Al(OH)}_{3}:\\\text{Moles of (CH$_{3}$COO)$_{3}$Al } =\text{3.526 Al(OH)}_{3} \times \dfrac{\text{1 mol (CH$_{3}$COO)$_{3}$Al }}{\text{1 mol Al(OH)}_{3}}\\\\= \text{3.526 mol (CH$_{3}$COO)$_{3}$Al}$

$\text{Acetic acid is the $\textbf{limiting reactant}$ because it gives fewer moles of} \\\text{(CH$_{3}$COO)$_{3}$Al. All $\textbf{125 g}$ will react.}$

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3 years ago