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Pani-rosa [81]
3 years ago
14

An unknown compound contains 75.69% carbon, 8.80% hydrogen, and 15.51% oxygen by mass. A mass spectrometry analysis reveals that

the molecular weight is 412 g/mol. What is the molecular formula of this compound?
Chemistry
1 answer:
tiny-mole [99]3 years ago
4 0

Answer:

Molecular formula is C₂₆H₃₆O₄

Explanation:

The compound is 75.69 % C, 8.80 % H and 15.51 % O. This data means, that in 100 g of compound we have 75.69 g, 15.51 g and 8.80 g of, C, O and H, respectively. We know the molar mass of the compound, so we can work to solve the moles of each element.

In 100 g of compound we have 75.69 g C, 15.51 g O and 8.80 g H

In 412 g of compound we would have:

(412 . 75.69) / 100 = 311.8 of C

(412 . 15.51) / 100 = 63.9 g of O

(412 . 8.80) / 100 = 36.2 g of H

Now, we can determine the moles of each, that are contained in 1 mol of compound.

312 g / 12 g/mol 26 C

64 g / 16 g/mol = 4 O

36 g / 1 g/mol = 36 H

Molecular formula is C₂₆H₃₆O₄

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Calculate either [ H 3 O + ] or [ OH − ] for each of the solutions.
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Answer: Solution A : [H_3O^+]=0.300\times 10^{-7}M

Solution B : [OH^-]=0.107\times 10^{-5}M

Solution C : [OH^-]=0.177\times 10^{-10}M

Explanation:

pH or pOH is the measure of acidity or alkalinity of a solution.

pH is calculated by taking negative logarithm of hydrogen ion concentration and pOH is calculated by taking negative logarithm of hydroxide ion concentration.

pH=-\log[H_3O^+]

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[H_3O^+][OH^-]=10^{-14}

a. Solution A: [OH^-]=3.33\times 10^{-7}M

[H_3O^+]=\frac{10^{-14}}{3.33\times 10^{-7}}=0.300\times 10^{-7}M

b. Solution B : [H_3O^+]=9.33\times 10^{-9}M

[OH^-]=\frac{10^{-14}}{9.33\times 10^{-9}}=0.107\times 10^{-5}M

c. Solution C : [H_3O^+]=5.65\times 10^{-4}M

[OH^-]=\frac{10^{-14}}{5.65\times 10^{-4}}=0.177\times 10^{-10}M

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