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storchak [24]
3 years ago
10

If a compound has a composition of 82% nitrogen and 18% hydrogen, what is the empirical formula for this compound

Chemistry
1 answer:
Damm [24]3 years ago
3 0

Answer: The empirical formula for the given compound is NH_3

Explanation : Given,

Percentage of H = 18 %

Percentage of N = 82 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of H = 18 g

Mass of N = 82 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{18g}{1g/mole}=18moles

Moles of Nitrogen = \frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{82g}{14g/mole}=5.8moles

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 5.8 moles.

For Hydrogen  = \frac{18}{5.8}=3.10\approx 3

For Nitrogen = \frac{5.8}{5.8}=1

Step 3: Taking the mole ratio as their subscripts.

The ratio of H : N = 3 : 1

Hence, the empirical formula for the given compound is NH_3

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[CH₃COO⁻] + [CH₃COOH] = 0.100M <em>(2)</em>

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1.820 = 0.100M - [CH₃COOH] / [CH₃COOH]

1.820[CH₃COOH] = 0.100M - [CH₃COOH]

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5.40 mL of a 0.490 M HCl are:

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HCl reacts with CH₃COO⁻ thus:

HCl + CH₃COO⁻ → CH₃COOH

After reaction, moles of CH₃COO⁻ are:

0.0101 moles - 2.646x10⁻³ moles = <em>7.429x10⁻³ moles of CH₃COO⁻</em>

<em />

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