Calculating Empirical Formulas with Percent Composition

Summary

This video tutorial walks viewers through solving a percent composition problem to determine the empirical formula of a compound. The instructor demonstrates a step-by-step approach, starting by assuming a 100-gram sample to convert percentages into grams, then converting grams to moles, and finding the mole ratio of elements. The process of adjusting fractional ratios to whole numbers is explained, leading to the final empirical formula. The video emphasizes careful calculations, using sufficient decimal places, and provides practical tips for translating percentages into a chemical formula, making it accessible for learners seeking to master empirical formula problems.

Takeaways

• 😀 Percent composition problems can be solved using the same method as other empirical formula problems.
• 😀 Assuming a 100-gram sample simplifies the conversion from percent to grams.
• 😀 Convert the mass of each element to moles using their molar masses.
• 😀 Molar mass of manganese (Mn) is approximately 54.94 g/mol and oxygen (O) is 16.00 g/mol.
• 😀 To find the mole ratio, divide the number of moles of each element by the smallest mole value.
• 😀 Initial mole ratios may include decimals, which need to be converted to whole numbers.
• 😀 Multiply the mole ratios by a common factor to eliminate decimals and achieve a whole number ratio.
• 😀 The final empirical formula uses the whole number mole ratio of elements.
• 😀 For the example in the script, the mole ratio of Mn to O is 2:3, giving the empirical formula Mn₂O₃.
• 😀 Significant figures are not critical in empirical formula calculations; focus on accurate mole ratios.
• 😀 This method is generalizable and works regardless of the sample size.

Q & A

What is the first step in solving a percent composition problem to find the empirical formula?

The first step is to assume a 100 gram sample. This allows the percentages of each element to be directly converted into grams, simplifying further calculations.

Why is it valid to assume a 100 gram sample when calculating empirical formulas?

It is valid because the chemical formula is independent of the sample size. Whether you have a small or large sample, the mole ratio between elements remains the same.

How do you convert grams of an element to moles?

Divide the mass of the element in grams by its molar mass (grams per mole). For example, moles of manganese = 69.6 g ÷ 54.94 g/mol ≈ 1.266 mol.

What is the next step after converting grams to moles for all elements?

Determine the mole ratio by dividing the number of moles of each element by the smallest number of moles among the elements.

What is the mole ratio of manganese to oxygen in the example given?

The initial mole ratio is approximately 1 mole of manganese to 1.5 moles of oxygen.

How do you handle a mole ratio that contains a decimal?

Multiply all ratios by a common factor to convert them to whole numbers. In this case, multiplying by 2 gives a whole number ratio of 2 manganese to 3 oxygen.

Why can't an empirical formula contain fractional subscripts?

Chemical formulas must have whole numbers of atoms because you cannot have a fraction of an atom in a molecule. Whole numbers represent actual atom counts.

What is the empirical formula of the compound in the example?

The empirical formula is Mn₂O₃.

How many decimal places should you consider when calculating moles?

It is advisable to keep a reasonable number of decimal places (e.g., 4) to maintain accuracy, but strict significant figures are less critical since the final answer is a whole-number formula.

What is the general process for solving percent composition problems?

1. Assume a 100 g sample. 2. Convert grams to moles using molar mass. 3. Divide each mole value by the smallest mole value to find the ratio. 4. Multiply to eliminate fractions and get whole numbers. 5. Write the empirical formula.

Why is it important to check your ratio by multiplying to get whole numbers?

Because the empirical formula requires integer subscripts, checking ensures that decimal ratios are correctly converted to the simplest whole-number ratio representing actual atom counts.