Question

An organic compound contains \(49.3 \%\) carbon, \(6.84 \%\) hydrogen and its vapour density is 73. Molecular formula of the compound is: (a) \(\mathrm{C}_{3} \mathrm{H}_{10} \mathrm{O}_{2}\) (b) \(\mathrm{C}_{3} \mathrm{H}_{5} \mathrm{O}_{2}\) (c) \(\mathrm{C}_{6} \mathrm{H}_{10} \mathrm{O}_{4}\) (d) \(\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}_{2}\)

Short answer

The molecular formula is \(\mathrm{C}_{6} \mathrm{H}_{10} \mathrm{O}_{4}\) (option c).

Step-by-step solution

Calculate the Mass of Oxygen

Since the compound is composed of carbon, hydrogen, and oxygen, and we know the percentages of carbon and hydrogen, we can calculate the percentage of oxygen. This is done by subtracting the sum of the percentages of carbon and hydrogen from 100.\[\text{Percentage of Oxygen} = 100\% - (49.3\% + 6.84\%) = 43.86\%\]

Find the Empirical Formula

To find the empirical formula, first, convert the percentages into grams assuming a 100 g sample. Then, divide each by the respective atomic masses of carbon (12 g/mol), hydrogen (1 g/mol), and oxygen (16 g/mol) to find the number of moles.\[\text{Moles of C} = \frac{49.3}{12} = 4.108\]\[\text{Moles of H} = \frac{6.84}{1} = 6.84\]\[\text{Moles of O} = \frac{43.86}{16} = 2.741\]Next, divide each by the smallest number of moles:\[\text{C: } \frac{4.108}{2.741} \approx 1.5 \\text{H: } \frac{6.84}{2.741} \approx 2.5 \\text{O: } \frac{2.741}{2.741} = 1\]The ratio of C:H:O is 1.5:2.5:1, which isn't whole numbers, so multiply all by 2 to achieve whole numbers: The empirical formula is \(\text{C}_3 \text{H}_5 \text{O}_2\).

Determine the Molecular Formula

The molecular weight can be determined using vapor density (which is given as 73). Molecular weight is twice the vapor density:\[\text{Molecular Weight} = 2 \times 73 = 146\]Compare this with the molar mass of the empirical formula, \(\text{C}_3 \text{H}_5 \text{O}_2\), which is calculated as follows:\[12 \times 3 + 1 \times 5 + 16 \times 2 = 73\]Since the empirical formula weight (73 Da) is half the molecular weight (146 Da), the molecular formula is twice the empirical formula. Thus the molecular formula is \(\text{C}_6 \text{H}_{10} \text{O}_4\).

Conclusion

Comparing our deduction with the given options, the molecular formula \(\text{C}_6 \text{H}_{10} \text{O}_4\) matches option (c). Thus, the correct answer is option (c).

Key concepts

Empirical Formula Calculation

The empirical formula is the simplest whole number ratio of atoms in a compound. It gives insight into the relative quantity of each type of atom present. To calculate an empirical formula, you start with the percentage composition of the compound. In this scenario, you assume a 100 g sample. This way, percentages seamlessly transform into grams. For example, 49.3% carbon becomes 49.3 g of carbon. Afterwards, convert grams to moles using atomic masses: carbon (12 g/mol), hydrogen (1 g/mol), and oxygen (16 g/mol). It involves dividing each element's gram by its atomic mass, producing the number of moles for each element. To simplify to a whole number ratio, divide each mole value by the smallest mole number calculated. If the results are not whole numbers, multiply each by a factor to reach whole numbers. For instance, if the ratio is 1.5, 2.5, and 1 for C, H, and O respectively, multiplying through by 2 gives C: 3, H: 5, O: 2. Thus, the empirical formula is \( \mathrm{C}_3 \mathrm{H}_5 \mathrm{O}_2 \).

Vapour Density and Molecular Weight

Vapour density provides a bridge to deducing molecular weights, especially for gaseous substances. The concept of vapour density stems from the formula:\[ \text{Vapour Density} = \frac{\text{Molar Mass}}{2} \]This relationship is key: doubling the vapour density gives the molar mass of the molecule. In our problem, the vapour density is 73, leading to a molecular weight calculated as:\[ 2 \times 73 = 146 \]With this information in hand, we match it up against the empirical formula's molar mass. The empirical formula \( \mathrm{C}_3 \mathrm{H}_5 \mathrm{O}_2 \) has a molar mass of 73 g/mol. Recognizing that this is half the molecular weight, we conclude the true molecular formula has twice the atoms for each element in the empirical formula, totaling \( \mathrm{C}_6 \mathrm{H}_{10} \mathrm{O}_4 \).

Organic Chemistry

Organic chemistry revolves around carbon. It explores the boundless forms and functions of carbon-based compounds. The problem discussed here is an organic compound made up of carbon, hydrogen, and oxygen. These elements frequently appear in organic compounds due to carbon's ability to form stable, varied structures. Understanding the ratios of these elements is vital in determining a compound's empirical and molecular formulas. Carbon's versatility allows the creation of various compounds with different physical and chemical properties. Knowledge of vapour density aids organic chemists in determining molecular weights, ensuring a comprehensive understanding of the molecules they work with. Organic compounds can hold functional groups altering their behavior, and understanding their formulas aids in predicting reactivity and interaction.

Stoichiometry in Chemistry

Stoichiometry focuses on the quantitative aspects of chemical reactions and compounds, ensuring that equations are balanced with proper proportions. The principles of stoichiometry are crucial in calculating both empirical formulas and molecular formulas. Starting from the percentages of elements, stoichiometry guides you to convert these into moles, then to a clear ratio for the empirical formula. It's an orderly process, emphasizing consistency and balance. For calculating the molecular formula, stoichiometry integrates the calculated molecular weight versus the known molar mass of the empirical formula. It involves proportionately scaling up the empirical formula to match the molecular weight. This demonstrates the core stoichiometric idea: conservation of particles and energy within reactions and compounds.