Question

A compound of carbon and hydrogen consists of \(94.34 \% \mathrm{C}\) and \(5.66 \% \mathrm{H},\) by mass. The molecular mass of the compound is found to be 178 u. What is its molecular formula?

Short answer

The molecular formula for the given carbon and hydrogen compound is \(C12H18\).

Step-by-step solution

Convert the percentage composition into grams

Since percentage is based out of 100, for simplicity we can consider we have 100 g of the compound. Therefore, if \(94.34 \% \mathrm{C}\) by mass, assume 94.34 g of Carbon. If \(5.66 \% \mathrm{H}\) then assume 5.66 g of Hydrogen.

Calculate the ratio of moles of Carbon and Hydrogen to find the empirical formula

Now, calculate the number of moles for each element. The number of moles = Mass of the element / molar mass of the element. Carbon: 94.34 g / 12.01 g/mol = 7.85 mol Carbon. Hydrogen: 5.66 g / 1.008 g/mol = 5.61 mol Hydrogen. After this, divide each of the calculated moles by the smallest quantity of moles calculated, to get the molar ratio. Carbon: 7.85 mol / 5.61 = 1.40 mol. Hydrogen: 5.61 mol / 5.61 = 1 mol. So, the empirical formula is \(CH_{1.4}\). But we only use integers in subscripts, and 1.4 is very close to 1.5, which when multiplied by 2 gives a whole number. Therefore, Multiply all the subscripts by 2 to convert them to whole numbers. This gives the Empirical formula as \(C2H3\).

Determine the molecular formula

The molecular formula is a multiple of the empirical formula. The molar mass of the empirical formula \(C2H3\) is 29 g/mol. Given the molecular mass is 178 u. Therefore, multiple = Molecular mass / Empirical formula mass = 178/29 = 6.13 ≈ 6. So, the molecular formula = 6 * Empirical formula = \(C12H18\).

Key concepts

Empirical Formula

An empirical formula represents the simplest whole-number ratio of atoms of each element in a compound. To determine the empirical formula, we start by converting the percentage composition of each element into grams. Assuming 100 grams of the compound simplifies calculations, as it allows us to directly equate percentages to grams.
For instance, in our original exercise, the compound consists of 94.34% carbon and 5.66% hydrogen. This translates to 94.34 grams of carbon and 5.66 grams of hydrogen. Next, we convert these masses into moles by dividing by the atomic masses: 12.01 g/mol for carbon and 1.008 g/mol for hydrogen.
Calculating gives us 7.85 moles of carbon and 5.61 moles of hydrogen. To find the simplest ratio, we divide by the smallest number of moles among the elements, here being 5.61. This yields a ratio of about 1.4:1 for carbon to hydrogen. Since we need whole numbers, we multiply the ratio by 2, resulting in the empirical formula \( C_2H_3 \).

Percentage Composition

Percentage composition is crucial for calculating an empirical formula. It informs us of the mass percentage of each element present in a compound.
For example, in the given exercise, the compound includes 94.34% of carbon and 5.66% of hydrogen. These values indicate how much of each element would be present in a 100-gram sample, as percentages effectively represent parts per hundred.
Given this composition, one can further deduce how to divide the compound into smaller parts, by converting these percentages into grams out of a standardized amount. This is essential for progressing in calculations necessary for deriving the empirical formula, providing a basis for establishing the simplest ratios between the elements involved.

Molar Mass Calculation

Molar mass calculation is pivotal in determining the molecular formula from an empirical formula. Once the empirical formula is known, calculating its molar mass involves the summation of the atomic masses of all constituent atoms.
In the given example, the empirical formula \( C_2H_3 \) was derived with its molar mass calculated as 29 g/mol. The actual molecular formula requires matching this molar mass against the compound's determined molar mass, which is 178 u.
To find the multiple, divide the molecular mass by the empirical formula mass: \( \frac{178}{29} \approx 6 \). This indicates that the empirical formula must be multiplied by 6, leading to the molecular formula \( C_{12}H_{18} \). This process ensures that the molecular formula accounts for the actual mass of the compound, providing a complete and accurate description of its composition.