Question

Determine the empirical and molecular formulas of each of the following substances: (a) Styrene, a compound substance used to make Styrofoam \(^{\circledast}\) 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 \(35.51 \%\) C, \(4.77 \%\) H, \(37.85 \%\) O, \(8.29 \% \mathrm{~N},\) and \(13.60 \% \mathrm{Na},\) and has a molar mass of \(169 \mathrm{~g} / \mathrm{mol}\)

Short answer

The molecular formulas for the given substances are: a) Styrene: C8H8 b) Caffeine: C8H10N4O2 c) Monosodium glutamate (MSG): CH2O4NNa

Step-by-step solution

Convert percentage composition to grams

We will use a 100g sample of styrene to make the calculations easier. Thus we have 92.3g of C and 7.7g of H.

Convert grams to moles

Use the molar masses of C and H to convert the masses into moles: moles of C = \(\frac{92.3g}{12.01g/mol}\), moles of H = \(\frac{7.7g}{1.01g/mol}\).

Calculate the empirical formula

Divide each value by the smallest number of moles and round to the nearest whole: \(\frac{moles \ of \ C}{moles \ of \ H} \approx 1\). Hence, the empirical formula is CH.

Determine the molecular formula

Using the empirical formula CH and the molar mass of 104 g/mol, we determine the molecular formula by multiplying the empirical formula by the appropriate factor: molecular formula = (empirical formula) x n. In this case, since the molar mass of CH is approximately 13 g/mol, we have (13 g/mol) x n = 104 g/mol → n = 8. Therefore, the molecular formula of styrene is C8H8. b) Caffeine

Convert percentage composition to grams

Using a 100g sample, we have 49.5g of C, 5.15g of H, 28.9g of N, and 16.5g of O.

Convert grams to moles

Find moles for each element: moles of C = \(\frac{49.5g}{12.01g/mol}\), moles of H = \(\frac{5.15g}{1.01g/mol}\), moles of N = \(\frac{28.9g}{14.01g/mol}\), moles of O = \(\frac{16.5g}{16.00g/mol}\).

Calculate the empirical formula

Divide each value by the smallest number of moles and round to the nearest whole: \(\frac{moles \ of \ C}{moles \ of \ H} \approx 4\), \(\frac{moles \ of \ N}{moles \ of \ H} \approx 2\), \(\frac{moles \ of \ O}{moles \ of \ H} \approx 1\). Hence, the empirical formula is C4H5N2O.

Determine the molecular formula

The molar mass of the empirical formula C4H5N2O is approximately 97 g/mol. Since the molar mass of caffeine is 195 g/mol, we have (97 g/mol) x n = 195 g/mol → n = 2. The molecular formula for caffeine is C8H10N4O2. c) Monosodium glutamate (MSG)

Convert percentage composition to grams

Using a 100g sample, we have 35.51g of C, 4.77g of H, 37.85g of O, 8.29g of N, and 13.60g of Na.

Convert grams to moles

Find moles for each element: moles of C = \(\frac{35.51g}{12.01g/mol}\), moles of H = \(\frac{4.77g}{1.01g/mol}\), moles of O = \(\frac{37.85g}{16.00g/mol}\), moles of N = \(\frac{8.29g}{14.01g/mol}\), moles of Na = \(\frac{13.6g}{22.99g/mol}\).

Calculate the empirical formula

Divide each value by the smallest number of moles and round to the nearest whole: \(\frac{moles \ of \ C}{moles \ of \ Na} \approx 1\), \(\frac{moles \ of \ H}{moles \ of \ Na} \approx 2\), \(\frac{moles \ of \ O}{moles \ of \ Na} \approx 4\), \(\frac{moles \ of \ N}{moles \ of \ Na} \approx 1\). Hence, the empirical formula for MSG is C1H2O4N1Na1 or simplified to CH2O4NNa.

Determine the molecular formula

The molar mass of the empirical formula CH2O4NNa is approximately 169 g/mol, which is the same as the given molar mass of MSG, indicating that the empirical formula and the molecular formula are the same. Therefore, the molecular formula for MSG is CH2O4NNa.

Key concepts

Percentage Composition

Understanding the percentage composition of a chemical compound is crucial as it gives an insight into the relative amounts of each element within that compound. For instance, suppose a student wants to determine the empirical and molecular formulas for styrene, which is reported to be composed of 92.3% carbon (C) and 7.7% hydrogen (H) by mass.

To begin with, we consider a 100g sample, which is a standard approach to simplify the math. This would mean the actual masses of the elements are 92.3g of carbon and 7.7g of hydrogen. The percentage composition directly translates to mass composition in grams when we use 100g of the substance, making the subsequent calculations for molar mass and moles conversion more straightforward.

Molar Mass Calculation

Molar mass is an essential property used to convert between the mass of a substance and the amount in moles. The molar mass of an element is its atomic weight in grams per mole, which can be found on the periodic table. Taking the element carbon as an example, it has a molar mass of 12.01 g/mol.

For compounds, the molar mass is the sum of the molar masses of all the atoms in one mole of the compound. During the analysis of chemical compounds like styrene, caffeine, or MSG, calculating the molar masses of these compounds helps in determining their molecular formulas; it acts as the piece connecting empirical formulas to molecular formulas.

Moles Conversion

Once the mass composition is known, the next step is to convert this mass into moles because chemical reactions are discussed in terms of moles rather than grams. This provides a common unit that relates a compound's mass to its number of atoms or molecules.

To convert the mass of an element to moles, the gram amount is divided by its molar mass. For instance, when calculating the moles of carbon in styrene, 92.3g (from the 100g sample) is divided by carbon's molar mass of 12.01 g/mol. This simple technique is used to balance chemical equations and calculate the empirical formula of a compound.

Chemical Compound Analysis

Chemical compound analysis involves the systematic steps taken to determine a compound's empirical and molecular formulas. This process starts with calculating the percentage composition, followed by converting this percentage into moles using the molar mass.

After obtaining moles for each element present, we find the simplest whole number ratio of the moles of elements to calculate the empirical formula. Finally, the molecular formula is deduced using the empirical formula and the compound's molar mass. This step-by-step approach reduces errors and ensures a clear path to understanding the composition of the substance under examination.