Find the empirical formula of the compound, given the relative atomic mass of silicon (Si) = 28 and oxygen (O) = 16. A compound contains 46.7% of silicon (Si) and 53.3% oxygen (O) by mass.

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Asked on March 26, 2026Chemistry

This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.

Accepted Answer · 2026-03-26T12:38:43.997Z

Q9. A compound contains 46.7% of silicon (Si) and 53.3% oxygen (O) by mass. Find the empirical formula of the compound, given the relative atomic mass of silicon (Si) = 28 and oxygen (O) = 16. Step 1: Assume a 100 g sample. This means we have 46.7 g of Si and 53.3 g of O. Step 2: Convert mass to moles for each element. Moles of Si = Mass of SiRelative atomic mass of Si = 46.7 g28 g/mol ≈ 1.6679 mol \\ Moles of O = Mass of ORelative atomic mass of O = 53.3 g16 g/mol ≈ 3.3313 mol Step 3: Find the simplest whole number ratio of moles. Divide each number of moles by the smallest number of moles (1.6679 mol). Ratio of Si = 1.6679 mol1.6679 mol = 1 \\ Ratio of O = 3.3313 mol1.6679 mol ≈ 1.997 ≈ 2 The simplest whole number ratio of Si:O is 1:2. Step 4: Write the empirical formula. The empirical formula is SiO_2. --- Q10. Caffeine is a compound found in coffee and tea. The percentage composition of caffeine is 49.1% carbon, 5.1% hydrogen, 16.5% oxygen and 28.9% nitrogen. The relative molecular mass of caffeine is 195. Find the molecular formula of caffeine, given the atomic mass of carbon (C) = 12, hydrogen (H) = 1, oxygen (O) = 16, and nitrogen (N) = 14. Step 1: Assume a 100 g sample and convert percentages to masses. Mass of C = 49.1 g Mass of H = 5.1 g Mass of O = 16.5 g Mass of N = 28.9 g Step 2: Convert mass to moles for each element. Moles of C = 49.1 g12 g/mol ≈ 4.0917 mol \\ Moles of H = 5.1 g1 g/mol = 5.1 mol \\ Moles of O = 16.5 g16 g/mol ≈ 1.0313 mol \\ Moles of N = 28.9 g14 g/mol ≈ 2.0643 mol Step 3: Find the simplest whole number ratio of moles. Divide each number of moles by the smallest number of moles (1.0313 mol for O). Ratio of C = 4.0917 mol1.0313 mol ≈ 3.967 ≈ 4 \\ Ratio of H = 5.1 mol1.0313 mol ≈ 4.945 ≈ 5 \\ Ratio of O = 1.0313 mol1.0313 mol = 1 \\ Ratio of N = 2.0643 mol1.0313 mol ≈ 2.001 ≈ 2 The simplest whole number ratio of C:H:O:N is 4:5:1:2. Step 4: Write the empirical formula. The empirical formula is C_4H_5ON_2. Step 5: Calculate the empirical formula mass. Empirical formula mass = (4 × 12) + (5 × 1) + (1 × 16) + (2 × 14) \\ = 48 + 5 + 16 + 28 \\ = 97 g/mol Step 6: Determine the factor 'n'. Given relative molecular mass = 195. n = Relative molecular massEmpirical formula mass = (195)/(97) ≈ 2.01 ≈ 2 Step 7: Write the molecular formula. Multiply the subscripts in the empirical formula by n=2. Molecular formula = (C_4H_5ON_2)_2 = C_8H_10O_2N_4

Accepted Answer · 2026-03-26T12:38:43.997Z

Q9. A compound contains 46.7% of silicon (Si) and 53.3% oxygen (O) by mass. Find the empirical formula of the compound, given the relative atomic mass of silicon (Si) = 28 and oxygen (O) = 16. Step 1: Assume a 100 g sample. This means we have 46.7 g of Si and 53.3 g of O. Step 2: Convert mass to moles for each element. Moles of Si = Mass of SiRelative atomic mass of Si = 46.7 g28 g/mol ≈ 1.6679 mol \\ Moles of O = Mass of ORelative atomic mass of O = 53.3 g16 g/mol ≈ 3.3313 mol Step 3: Find the simplest whole number ratio of moles. Divide each number of moles by the smallest number of moles (1.6679 mol). Ratio of Si = 1.6679 mol1.6679 mol = 1 \\ Ratio of O = 3.3313 mol1.6679 mol ≈ 1.997 ≈ 2 The simplest whole number ratio of Si:O is 1:2. Step 4: Write the empirical formula. The empirical formula is SiO_2. --- Q10. Caffeine is a compound found in coffee and tea. The percentage composition of caffeine is 49.1% carbon, 5.1% hydrogen, 16.5% oxygen and 28.9% nitrogen. The relative molecular mass of caffeine is 195. Find the molecular formula of caffeine, given the atomic mass of carbon (C) = 12, hydrogen (H) = 1, oxygen (O) = 16, and nitrogen (N) = 14. Step 1: Assume a 100 g sample and convert percentages to masses. Mass of C = 49.1 g Mass of H = 5.1 g Mass of O = 16.5 g Mass of N = 28.9 g Step 2: Convert mass to moles for each element. Moles of C = 49.1 g12 g/mol ≈ 4.0917 mol \\ Moles of H = 5.1 g1 g/mol = 5.1 mol \\ Moles of O = 16.5 g16 g/mol ≈ 1.0313 mol \\ Moles of N = 28.9 g14 g/mol ≈ 2.0643 mol Step 3: Find the simplest whole number ratio of moles. Divide each number of moles by the smallest number of moles (1.0313 mol for O). Ratio of C = 4.0917 mol1.0313 mol ≈ 3.967 ≈ 4 \\ Ratio of H = 5.1 mol1.0313 mol ≈ 4.945 ≈ 5 \\ Ratio of O = 1.0313 mol1.0313 mol = 1 \\ Ratio of N = 2.0643 mol1.0313 mol ≈ 2.001 ≈ 2 The simplest whole number ratio of C:H:O:N is 4:5:1:2. Step 4: Write the empirical formula. The empirical formula is C_4H_5ON_2. Step 5: Calculate the empirical formula mass. Empirical formula mass = (4 × 12) + (5 × 1) + (1 × 16) + (2 × 14) \\ = 48 + 5 + 16 + 28 \\ = 97 g/mol Step 6: Determine the factor 'n'. Given relative molecular mass = 195. n = Relative molecular massEmpirical formula mass = (195)/(97) ≈ 2.01 ≈ 2 Step 7: Write the molecular formula. Multiply the subscripts in the empirical formula by n=2. Molecular formula = (C_4H_5ON_2)_2 = C_8H_10O_2N_4

Find the empirical formula of the compound, given the relative atomic mass of silicon (Si) = 28 and oxygen (O) = 16. A compound contains 46.7% of silicon (Si) and 53.3% oxygen (O) by mass.

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Find the empirical formula of the compound, given the relative atomic mass of silicon (Si) = 28 and oxygen (O) = 16. A compound contains 46.7% of silicon (Si) and 53.3% oxygen (O) by mass.

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