Determine the empirical and molecular formulas of each of the following substances: (a) Styrene, a compound used to make Styrofoam \(^{*}\) cups and insulation, contains \(92.3 \% \mathrm{C}\) and \(7.7 \% \mathrm{H}\) by mass and has a molar mass of \(104 \mathrm{~g} / \mathrm{mol}\). (b) Caffeine, a stimulant found in coffee, contains \(49.5 \% \mathrm{C}\), \(5.15 \% \mathrm{H}, 28.9 \% \mathrm{~N},\) and \(16.5 \% \mathrm{O}\) by mass and has a molar mass of \(195 \mathrm{~g} / \mathrm{mol}\) (c) Monosodium glutamate (MSG), a flavor enhancer in certain foods, contains \(35.51 \% \mathrm{C}, 4.77 \% \mathrm{H}, 37.85 \% \mathrm{O},\) $8.29 \% \mathrm{~N},\( and \)13.60 \% \mathrm{Na},\( and has a molar mass of \)169 \mathrm{~g} / \mathrm{mol} .$

Step by step solution

01

Convert mass percentages to grams

Since the mass percentages are given, assume a 100g sample. This means we have: - 92.3g of Carbon (C) - 7.7g of Hydrogen (H)

02

Convert grams to moles

Using the molar masses, convert the grams to moles: - Moles of C: \( \frac{92.3g}{12.01g/mol} = 7.69 \, mol \) - Moles of H: \( \frac{7.7g}{1.01g/mol} = 7.62 \, mol \)

03

Find the mole ratio

Divide the moles by the smallest number of moles: - Mole ratio of C: \( \frac{7.69}{7.62} = 1.01\) - Mole ratio of H: \( \frac{7.62}{7.62} = 1\)

04

Obtain whole-number subscripts

The mole ratios are already nearly whole numbers, so no need to multiply. The subscripts are: - C: 1 - H: 1

05

Determine the empirical formula

The empirical formula of Styrene is CH.

06

Calculate the empirical formula mass and find the molecular formula

The empirical formula mass for CH is (12.01 + 1.01) = 13.02 g/mol. Since the molar mass is given as 104 g/mol, we can divide it by the empirical formula mass: - \( \frac{104}{13.02} = 8 \) Thus, the molecular formula is 8(CH) which is C8H8. #(b) Caffeine# Following the same steps as above:

07

Convert mass percentages to grams

Assume a 100g sample: - 49.5g of Carbon (C) - 5.15g of Hydrogen (H) - 28.9g of Nitrogen (N) - 16.5g of Oxygen (O)

08

Convert grams to moles

Using the molar masses, convert the grams to moles: - Moles of C: \( \frac{49.5g}{12.01g/mol} = 4.12 \, mol \) - Moles of H: \( \frac{5.15g}{1.01g/mol} = 5.1 \, mol \) - Moles of N: \( \frac{28.9g}{14.01g/mol} = 2.06 \, mol \) - Moles of O: \( \frac{16.5g}{16.00g/mol} = 1.03 \, mol \)

09

Find the mole ratio

Divide the moles by the smallest number of moles: - Mole ratio of C: \( \frac{4.12}{1.03} = 4 \) - Mole ratio of H: \( \frac{5.1}{1.03} = 5 \) - Mole ratio of N: \( \frac{2.06}{1.03} = 2 \) - Mole ratio of O: \( \frac{1.03}{1.03} = 1 \)

10

Determine the empirical formula

The empirical formula of Caffeine is C4H5N2O.

11

Calculate the empirical formula mass and find the molecular formula

The empirical formula mass for C4H5N2O is (4*12.01 + 5*1.01 + 2*14.01 + 1*16.00) = 97.09 g/mol. Since the given molar mass is 195 g/mol, we can divide it by the empirical formula mass: - \( \frac{195}{97.09} = 2 \) Thus, the molecular formula of Caffeine is 2(C4H5N2O) which is C8H10N4O2. #(c) Monosodium glutamate (MSG)# Following the same steps as above:

12

Convert mass percentages to grams

Assume a 100g sample: - 35.51g of Carbon (C) - 4.77g of Hydrogen (H) - 37.85g of Oxygen (O) - 8.29g of Nitrogen (N) - 13.60g of Sodium (Na)

13

Convert grams to moles

Using the molar masses, convert the grams to moles: - Moles of C: \( \frac{35.51g}{12.01g/mol} = 2.96 \, mol \) - Moles of H: \( \frac{4.77g}{1.01g/mol} = 4.72 \, mol \) - Moles of O: \( \frac{37.85g}{16.00g/mol} = 2.37 \, mol \) - Moles of N: \( \frac{8.29g}{14.01g/mol} = 0.592 \, mol \) - Moles of Na: \( \frac{13.60g}{22.99g/mol} = 0.592 \, mol \)

14

Find the mole ratio

Divide the moles by the smallest number of moles: - Mole ratio of C: \( \frac{2.96}{0.592} = 5 \) - Mole ratio of H: \( \frac{4.72}{0.592} = 8 \) - Mole ratio of O: \( \frac{2.37}{0.592} = 4 \) - Mole ratio of N: \( \frac{0.592}{0.592} = 1 \) - Mole ratio of Na: \( \frac{0.592}{0.592} = 1 \)

15

Determine the empirical formula

The empirical formula of MSG is C5H8O4N1Na1, or C5H8O4NNa.

16

Calculate the empirical formula mass and find the molecular formula

The empirical formula mass for C5H8O4NNa is (5*12.01 + 8*1.01 + 4*16.00 + 14.01 + 22.99) = 169.00 g/mol. Since the given molar mass is 169 g/mol, the empirical formula mass and molar mass are equal. Thus, the molecular formula of Monosodium glutamate is C5H8O4NNa.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!