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Alika [10]
2 years ago
6

A sample of pure lithium chloride contains 16% lithium by mass. What is the % lithium by mass in a sample of pure lithium carbon

ate that has twice the mass of the first sample
Chemistry
1 answer:
Anna35 [415]2 years ago
3 0

Answer:

Percentage lithium by mass in Lithium carbonate sample = 19.0%

Explanation:

Atomic mass of lithium = 7.0 g; atomic mass of Chlorine = 35.5 g; atomic mass of carbon = 12.0 g; atomic mass of oxygen = 16.0 g

Molar mass of lithium chloride, LiCl = 7 + 35.5 = 42.5 g

Percentage by mass of lithium in LiCl = (7/42.5) * 100% = 16.4 % aproximately 16%

Molar mass of lithium carbonate, Li₂CO₃ = 7 * 2 + 12 + 16 * 3 =74.0 g

Percentage by mass of lithium in Li₂CO₃ = (14/74) * 100% = 18.9 % approximately 19%

Mass of Lithium carbonate sample = 2 * 42.5 = 85.0 g

mass of lithium in 85.0 g Li₂CO₃ = 19% * 85.0 g = 16.15 g

Percentage by mass of lithium in 85.0 g Li₂CO₃ = (16.15/85.0) * 100 % = 19.0%

Percentage lithium by mass in Lithium carbonate sample = 19.0%

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Answer:

The empirical formula is = C_4H_8O

The formula of Valproic acid = C_8H_{16}O_2

Explanation:

Mass of water obtained = 0.166 g

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Moles of H_2O = 0.166 g /18 g/mol = 0.00922 moles

2 moles of hydrogen atoms are present in 1 mole of water. So,

<u>Moles of H = 2 x 0.00922 = 0.01844 moles </u>

Molar mass of H atom = 1.008 g/mol

<u>Mass of H in molecule = 0.01844 x 1.008 = 0.018588 g </u>

Mass of carbon dioxide obtained = 0.403 g

Molar mass of carbon dioxide = 44.01 g/mol

Moles of CO_2 = 0.403 g  /44.01 g/mol = 0.009157 moles

1 mole of carbon atoms are present in 1 mole of carbon dioxide. So,

<u>Moles of C = 0.009157 moles </u>

Molar mass of C atom = 12.0107 g/mol

<u>Mass of C in molecule = 0.009157 x 12.0107 = 0.11 g </u>

<u>Given that the Valproic acid only contains hydrogen, oxygen and carbon. </u>So,

Mass of O in the sample = Total mass - Mass of C  - Mass of H

Mass of the sample = 0.165 g

<u>Mass of O in sample = 0.165 - 0.11 - 0.018588 = 0.036412 g  </u>

Molar mass of O = 15.999 g/mol

<u>Moles of O  = 0.036412  / 15.999  = 0.002276 moles</u>

<u></u>

<u>Taking the simplest ratio for H, O and C as: </u>

<u>0.01844 : 0.002276 : 0.009157</u>

<u> = 8 : 1 : 4</u>

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Molecular mass = n × Empirical mass

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Mass from the Empirical formula = 4×12 + 8×1 + 16= 72 g/mol

Molar mass = 144 g/mol

So,  

Molecular mass = n × Empirical mass

144 = n × 72

<u>⇒ n = 2</u>

The formula of Valproic acid = C_8H_{16}O_2

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