Question

A sample of a hydrocarbon (a compound consisting of only carbon and hydrogen contains \(2.59 \times 10^{23}\) atoms of hydrogen and is 17.3\(\%\) hydrogen by mass. If the molar mass of the hydrocarbon is between 55 and 65 g/mol, what amount (moles) of compound is present, and what is the mass of the sample?

Short answer

After determining the moles of hydrogen and carbon in the sample, we find the ratio between them, which helps us identify the empirical formula of the hydrocarbon (CmHn). We then calculate the empirical formula molar mass and use the given molar mass range (55-65 g/mol) to find the moles of the hydrocarbon. Finally, we calculate the mass of the sample using the moles of the compound and the given molar mass of the hydrocarbon.

Step-by-step solution

Determine hydrogen moles the from given hydrogen atoms

We are given the number of hydrogen atoms in the sample: \(2.59 \times 10^{23}\). To find the moles of hydrogen, we'll divide this number by Avogadro's number, \(6.022 \times 10^{23} \, \text{atoms/mol}\). Hydrogen moles = \(\frac{2.59 \times 10^{23} \, \text{atoms}}{6.022 \times 10^{23} \, \text{atoms/mol}}\)

Calculate the moles of carbon using percent hydrogen by mass

We are given that the sample is 17.3% hydrogen by mass. As a hydrocarbon only has carbon and hydrogen, we can assume that the rest of the mass is contributed by carbon (100% - 17.3% = 82.7%). Now we need to find the mass of hydrogen in the sample. As we know the moles of hydrogen, we can determine the mass of hydrogen using the molar mass of hydrogen, which is approximately 1 g/mol. Mass of hydrogen = Hydrogen moles × Molar mass of hydrogen = Hydrogen moles × 1 g/mol Now, let 'x' be the total mass of the sample: 17.3% × x = Mass of hydrogen Calculate the mass of carbon (82.7% of the sample) and divide it by the molar mass of carbon (12 g/mol) to find the moles of carbon. Carbon moles = \(\frac{\text{Mass of carbon (82.7%)}\times x}{\text{Molar mass of carbon (12 g/mol)}}\)

Determine the ratio between carbon and hydrogen atoms

Divide the moles of carbon by the moles of hydrogen to find the ratio between the two: \( \frac{\text{Carbon moles}}{\text{Hydrogen moles}} = \frac{C}{H}\)

Calculate the empirical formula

Now, using the ratio found in Step 3, we can identify the empirical formula of the hydrocarbon. Let's assume it is in the form CmHn, wherein we need to find the integers m and n.

Find the molar mass of the empirical formula

Calculate the molar mass of the empirical formula (CmHn) using the molar masses of carbon and hydrogen and the values of m and n. Empirical formula molar mass = (12 g/mol × m) + (1 g/mol × n)

Calculate the moles of the compound and the mass of the sample using the empirical formula

Using the molar mass range given (55-65 g/mol), divide the given molar mass of the hydrocarbon by the empirical formula's molar mass (found in Step 5) to find the moles of the compound, as follows: Moles of hydrocarbon = \(\frac{\text{Given molar mass}}{\text{Empirical formula molar mass}}\) Finally, we can find the mass of the sample using the calculated moles of the compound and the given molar mass of the hydrocarbon, as follows: Sample mass = Moles of hydrocarbon × Given molar mass

Key concepts

Empirical Formula

The empirical formula of a compound gives the simplest whole-number ratio of atoms of each element in the compound. In the case of hydrocarbons, the empirical formula represents the basic building block of the compound, consisting only of carbon and hydrogen atoms. To determine this ratio, we need to know the number of moles of each element present. This can be obtained from their respective masses and molar masses.
In our hydrocarbon sample, the procedure starts with understanding the proportion of hydrogen (17.3% by mass). Thus, 82.7% is carbon by mass. From this, we calculate the moles of each element, assuming a total mass of the sample (e.g., 100g for ease of calculation). The moles of hydrogen are derived from the number of hydrogen atoms divided by Avogadro's number. For carbon, it involves subtracting the hydrogen mass from the total mass and using carbon's molar mass (12 g/mol) to find carbon moles.
This ratio is crucial as it forms the basis for the empirical formula. The step-wise comparison of moles of hydrogen and carbon gives the simplest ratio, thereby determining the empirical formula CmHn for the hydrocarbon in question.

Molar Mass Calculation

Calculating the molar mass of a hydrocarbon involves using the found empirical formula. This is done by multiplying the number of carbons and hydrogens in the empirical formula by their respective molar masses (12 g/mol for carbon and 1 g/mol for hydrogen).
The empirical formula gives a theoretical molar mass, which can assist as a benchmark when comparing with the actual molar mass range—as indicated in the problem (55-65 g/mol). This comparison helps refine our understanding of the compound's composition, whether the empirical formula represents the true formula or a simplified version.
To see if the empirical formula matches the molar mass range given, we multiply the discrete atomic masses by the integers in the derived empirical formula. Any discrepancies between the calculated and given molar mass might suggest a need to scale the empirical formula by an integer factor (e.g., doubling the empirical formula if its molar mass is half of the actual observed range). This way, the empirical formulary is validated or revised to match the observed compound.

Atomic Composition Analysis

Atomic composition analysis helps in verifying the identities and quantities of each element present in a compound. For hydrocarbons, this analysis focuses on understanding how much hydrogen and carbon is present through their moles.
The first step in this process is quantifying hydrogen atoms in moles using Avogadro's number. This number reveals the relative amount of hydrogen compared to carbon, allowing for calculations of composition and supporting the development of the empirical formula.
A strategic element here is assuming a convenient total mass for calculations—often 100 grams—to simplify percentage conversion to mass calculation. This aids in aligning the percentage of each element contribution to total atomic masses, with weight percent figures turning directly into grams. From here, it divides for moles using known atomic masses, authenticating empirical ratios directly through quantifiable scientific measures, translating this analysis into actionable data needed for further formula and mass precision.