A compound whose empirical formula is XF3 consists of 65% F by mass. What is the atomic mass of X?

검증된 단계별 안내

Determine the mass percentage of element X in the compound by subtracting the mass percentage of F from 100%. This is calculated as 100% - 65% = 35%.

Assume a sample size of 100 grams for easy calculation. This means that in a 100 gram sample, 35 grams would be element X and 65 grams would be F.

Use the molar mass of F (fluorine), which is approximately 19 g/mol, to find the moles of F in the compound. Calculate this by dividing the mass of F by its molar mass: 65 grams \(\div\) 19 g/mol.

Since the empirical formula is XF3, there are three moles of F for every mole of X. Use the moles of F calculated in the previous step to find the moles of X by dividing the moles of F by 3.

Calculate the molar mass of X by dividing the mass of X (35 grams) by the moles of X calculated in the previous step. This will give you the atomic mass of X.

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주요 개념

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Empirical Formula

An empirical formula represents the simplest whole-number ratio of the elements in a compound. In this case, XF3 indicates that for every one atom of element X, there are three atoms of fluorine (F). Understanding empirical formulas is crucial for determining the composition and calculating the molar mass of compounds.

Mass Percent Composition

Mass percent composition is the percentage by mass of each element in a compound. It is calculated by dividing the mass of the element in one mole of the compound by the molar mass of the compound, then multiplying by 100. In this question, knowing that fluorine constitutes 65% of the compound's mass allows us to set up equations to find the atomic mass of X.

Molar Mass Calculation

Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). To find the atomic mass of element X, we can use the mass percent of fluorine and the empirical formula to establish a relationship between the total molar mass of the compound and the individual contributions of each element. This calculation is essential for solving the problem.