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Dominik [7]
3 years ago
13

Some of the formulas below could be either molecular or empirical formulas; however, some could only be molecular formulas. Whic

h of the following formulas must be molecular formulas? Select all that apply.
H2O2

CH

C4H8O4

C4H5N2O

NO2
Chemistry
2 answers:
kondor19780726 [428]3 years ago
3 0
H2O2 and C4H8O4 must be molecular formulas.
Evgen [1.6K]3 years ago
3 0

Answer: H_2O_2

C_4H_8O_4

Explanation:

Molecular formula is the chemical formula which depicts the actual number of atoms of each element present in the compound.  

Empirical formula is the simplest chemical formula which depicts the whole number of atoms of each element present in the compound.  

H_2O_2 must be a molecular formula as empirical formula will be OH

CH must be a empirical formula.

C_4H_8O_4 must be a molecular formula as the empirical formula will be CH_2O

C_4H_5N_2O can be a molecular formula or a empirical formula.

NO_2 can be a molecular formula or a empirical formula.

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Answer:

C. Dalton

Explanation:

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Upon combustion, a compound containing only carbon and hydrogen produces 2.67 g<img src="https://tex.z-dn.net/?f=CO_%7B2%7D" id=
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Answer:

\boxed{\text{CH$_{2}$}}

Explanation:

1. Calculate the mass of each element

\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}

2. Calculate the moles of each element

\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}

3. Calculate the molar ratios

Divide all moles by the smallest number of moles.

\text{C: } \dfrac{0.06067}{0.06067}= 1\\\\\text{H: } \dfrac{0.1221}{0.06067} = 2.015

4. Round the ratios to the nearest integer

C:H = 1:2

5. Write the empirical formula

The empirical formula is \boxed{\text{CH$_{2}$}}

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