Problem 120 - Chemistry The Molecular Science

Question

A hydrocarbon is \(90.0 \%\) carbon and \(10.0 \%\) hydrogen by mass; its molar mass is \(40.1 \mathrm{~g} / \mathrm{mol}\). The hydrocarbon reacts with bromine by an addition reaction. It also undergoes hydrogenation with a nickel catalyst; \(1.58 \mathrm{~g}\) of the hydrocarbon reacts with \(1.77 \mathrm{~L} \mathrm{H}_{2}\) (measured at STP). (a) From these data, derive the molecular formula of the hydrocarbon. (b) Write the structural formulas of two possible isomers. (c) For one of the isomers, use Lewis structures to write a balanced chemical equation for: (i) the bromination reaction; and (ii) the hydrogenation reaction.

Step-by-Step Solution

Answer

The molecular formula is \( \text{C}_3\text{H}_4 \). Possible isomers are propyne and propadiene. Example bromination: \( \text{CH}_2\text{=C}=\text{CH}_2 + \text{Br}_2 \rightarrow \text{CH}_2\text{Br}-\text{CHBr}=\text{CH}_2 \); hydrogenation: \( \text{CH}_2\text{=C}=\text{CH}_2 + 2\text{H}_2 \rightarrow \text{CH}_3\text{-CH}_3\text{-CH}_3 \).

Step 1: Determine Empirical Formula

The hydrocarbon is composed of 90.0% carbon and 10.0% hydrogen by mass. We assume 100 g of the substance to facilitate calculations. This gives us 90 g of carbon and 10 g of hydrogen.To find moles of each element:- Carbon: \( \text{moles of C} = \frac{90 \, g}{12.01 \, g/mol} \approx 7.49 \, mol \)- Hydrogen: \( \text{moles of H} = \frac{10 \, g}{1.008 \, g/mol} \approx 9.92 \, mol \)Next, divide moles by the smaller number to find the simplest ratio:- Ratio of C:H = \( \frac{7.49}{7.49} : \frac{9.92}{7.49} \approx 1 : 1.32 \)Since we need whole numbers, we multiply each number by 3 (common multiplier) to get:- Carbon : Hydrogen = 3 : 4Thus, empirical formula is \( \text{C}_3\text{H}_4 \).

Step 2: Determine Molecular Formula

The molar mass of the empirical formula \( \text{C}_3\text{H}_4 \) is:\[ \text{Molar mass of } \text{C}_3\text{H}_4 = 3 \times 12.01 + 4 \times 1.008 = 40.05 \, g/mol \]Since this matches closely with the given molar mass of \( 40.1 \, g/mol \), the empirical formula is also the molecular formula of the hydrocarbon, which is \( \text{C}_3\text{H}_4 \).

Step 3: Identify Possible Isomers

The hydrocarbon \( \text{C}_3\text{H}_4 \) could be linear or branched. Possible isomers include propyne and propadiene:1. Propyne: \( \text{CH}_3\text{-C≡C-H} \)2. Propadiene: \( \text{CH}_2\text{=C}=\text{CH}_2 \)

Step 4: Write Bromination Reaction for One Isomer

For propadiene (\( \text{CH}_2\text{=C}=\text{CH}_2 \)) reacting with bromine, the reaction is:\[ \text{CH}_2\text{=C}=\text{CH}_2 + \text{Br}_2 \rightarrow \text{CH}_2\text{Br}-\text{CHBr}=\text{CH}_2 \]

Step 5: Write Hydrogenation Reaction for One Isomer

For propadiene (\( \text{CH}_2\text{=C}=\text{CH}_2 \)) undergoing hydrogenation with hydrogen in the presence of a nickel catalyst:\[ \text{CH}_2\text{=C}=\text{CH}_2 + 2\text{H}_2 \rightarrow \text{CH}_3\text{-CH}_3\text{-CH}_3 \]This hydrocarbon converts to propane in the reaction.

Key Concepts

Empirical Formula

To determine the empirical formula, one must start by understanding what it signifies. It is the simplest whole-number ratio of atoms in a compound, giving insights into the basic proportions of elements involved, but not necessarily the actual numbers of atoms in a molecule.
In our example, we have a hydrocarbon made of 90% carbon and 10% hydrogen by mass. Assuming we have 100 grams of the hydrocarbon, we would have 90 grams of carbon and 10 grams of hydrogen. To convert these masses to moles:

• For carbon: divide the mass by the atomic mass: \( \frac{90}{12.01} \approx 7.49 \text{ mol} \)
• For hydrogen: divide the mass by the atomic mass: \( \frac{10}{1.008} \approx 9.92 \text{ mol} \)

To find the ratio, compare these mole values. Dividing both by the smallest number of moles gives a ratio which, when approximated to whole numbers, provides the empirical formula. Here, the ratio simplifies to a 3:4 ratio of carbon to hydrogen, leading to the empirical formula \( \text{C}_3\text{H}_4 \).

Isomers

Isomers are fascinating because they share the same molecular formula but have different structural arrangements. This means that, while they are composed of the same types and numbers of atoms, their shapes or alignments in space differ, resulting in varying properties.Consider the hydrocarbon \( \text{C}_3\text{H}_4 \). It can form different isomers such as propyne and propadiene:

• Propyne: Geometrically simple with a linear form, represented as \( \text{CH}_3\text{-}\text{C} \equiv \text{C}\text{-H} \).
• Propadiene: Occurs as a more complex structure with double bonds, written as \( \text{CH}_2 = \text{C} = \text{CH}_2 \).

These distinct structures show how simple changes in bonding and alignment can lead to entirely different chemicals even though they stem from the same molecular formula.

Addition Reaction

An addition reaction is a type of chemical reaction where atoms are added to a molecule, generally involving compounds that possess double or triple bonds.For our hydrocarbon \( \text{C}_3\text{H}_4 \), let's explore its addition reaction with bromine. In such a reaction, the double bond of propadiene (\( \text{CH}_2 = \text{C} = \text{CH}_2 \)) opens up, allowing bromine atoms (\( \text{Br}_2 \)) to attach to the former double-bonded carbons:

• The reaction is \( \text{CH}_2 = \text{C} = \text{CH}_2 + \text{Br}_2 \rightarrow \text{CH}_2\text{Br}-\text{CHBr}=\text{CH}_2 \).
• This reaction results in the incorporation of bromine, forming a new compound by adding across the carbon double bond.

This type of reaction exemplifies how unsaturated bonds can convert into saturated or partially saturated bonds by adding elements directly to the original compound structure.

Hydrogenation Reaction

Hydrogenation is an exciting reaction often encountered in organic chemistry, involving the addition of hydrogen (\( \text{H}_2 \)) to a compound, typically with a catalyst such as nickel.In our case with \( \text{C}_3\text{H}_4 \), hydrogenation occurs when the unsaturated propadiene is treated with hydrogen gas in the presence of a nickel catalyst:

• The reaction is \( \text{CH}_2 = \text{C} = \text{CH}_2 + 2\text{H}_2 \rightarrow \text{CH}_3-\text{CH}_2-\text{CH}_3 \).
• This process adds hydrogen across the double bonds, converting propadiene into propane, a fully saturated molecule with only single bonds.

Hydrogenation is not only a valuable industrial tool, creating more stable compounds, but also highlights the transformation of highly reactive unsaturated molecules into more stable saturated ones.