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abruzzese [7]
2 years ago
9

Combustion analysis of toluene, a common organic solvent, gives 7.03 mgmg of CO2 and 1.64 mgmg of H2O. If the compound contains

only carbon and hydrogen, what is its empirical formula?
Chemistry
1 answer:
Anika [276]2 years ago
6 0

Answer:

The empirical formula is = C_7H_8

Explanation:

Mass of carbon dioxide obtained = 7.03 mg

1 mg = 10⁻³ g

Molar mass of carbon dioxide = 44.01 g/mol

Moles of CO_2 = 7.03× 10⁻³ g  /44.01 g/mol = 0.1597×10⁻³ moles

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

Moles of C = 0.1597×10⁻³ moles

Mass of water obtained = 1.64 mg

Moles of H_2O = 1.64× 10⁻³ g /18 g/mol = 0.0911×10⁻³ moles

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

Moles of H = 2 x 0.0911 ×10⁻³  = 0.1822 ×10⁻³ moles

Taking the simplest ratio for H and C as:

0.1822 ×10⁻³ : 0.1597×10⁻³

 = 8 : 7

The empirical formula is = C_7H_8

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Answer:

pH = 11.95≈12

Explanation:

Remember  the reaction among aqueous acetic acid (CH_3COOH) and aqueous sodium hydroxide (NaOH)

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First step. Need to know how much moles of the substances are present

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Careful: we have to use the total volumen

Les us to calculate pH

pH= -log [H+]\\pH= -log \frac{K_w}{[OH-]} \\pH= 11.95

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