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nalin [4]
3 years ago
9

A 425.00-gram sample of a compound decomposes into 196.01 grams of carbon, 41.14 grams of hydrogen, 130.56 grams of oxygen, and

57.29 grams of silicon. Experiments have shown the compound has a molecular weight of 208.329. What is the molecular formula?
A.
C8H5O2Si
B.
C2H5OSi
C.
C4H10O2Si
D.
C8H20O4Si
Chemistry
1 answer:
Alenkasestr [34]3 years ago
3 0

The molecular formula is D. C_8H_20O_4Si.

<em>Step 1</em>.Calculate the <em>empirical formula </em>

a) Calculate the moles of each element

Moles of C= 196.01 g C × (1 mol C/12.01 g C) = 16.325 mol C  

Moles of H = 41.14 g H × (1 mol H/1.008 g H) = 40.813 mol H

Moles of O = 130.56 g O × (1 mol O/16.00 g O) = 8.1650 mol O

Moles of Si = 57.29 g Si × (1 mol Si/28.085 g Si) = 2.0399 mol Si

b) Calculate the molar ratio of each element

Divide each number by the smallest number of moles and round off to an integer

C:H:O:Si = 8.0027:20.008:4.0027:1 ≈ 8:20:4:1

c) Write the empirical formula

EF = C_8H_20O_4Si

<em>Step </em>2. Calculate the <em>molecular formula</em>

EF Mass = 208.33 u

MF mass = 208.329  u

MF = (EF)_n

n = MF Mass/EF Mass = 208.329 u/208.33 u = 1.0000 ≈ 1

MF = C_8H_20O_4Si

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Explanation:

We will need a balanced equation with masses, moles, and molar masses, so let’s gather all the information in one place.

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\rm  \text{Moles of CO}_{2} = \text{1.5 mol C$_{3}$H}_{8} \times \dfrac{\text{3 mol CO}_{2}}{\text{1 mol C$_{3}$H}_{8}} =\text{4.5 mol CO}_{2}

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\text{Mass of CO}_{2} = \text{4.5 mol CO}_{2}  \times \dfrac{\text{44.01 g CO}_{2}}{\text{1 mol CO$_{2}$}} = \textbf{200 g CO}_{\mathbf{2}}\\\text{The reaction will form $\large \boxed{\textbf{200 g CO}_{\mathbf{2}}}$}

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