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Hey I think I understand this a little but I need some help figuring out the answers to number 2,3,4,5,6,7,8,9and10 There are examples but I don't understand it that much

Hey I think I understand this a little but I need some help figuring out the answers to number 2,3,4,5,6,7,8,9and10 There are examples but I don't understand it that much

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Hey I think I understand this a little but I need some help figuring out the answers to number 2,3,4,5,6,7,8,9and10 There are examples but I don't understand it that much
Chem Activity 055 Determining Empirical Formulas What is the empirical formula (lowest whole number ratio) of the compounds below? Example 1: 11% H, 88% o Percent grams divide by molar mass = Moles Divide each by smaller 1 g/mol 16g/mol *Assume that 100 % is 100 grams = 11 moles/5.5 = 2 = 5.5 moles/5.5 = 1 mole Ratio *11 g H / 88% O = *88 g O / 11% H = 2 parts H to 1 part O so Example 2: 30% O, 70% Fe Percent grams divide by molar mass = Moles 30% O, so 30 g0 / 16 g/mol = 1.875/1.25 = 1.5 moles x 2 = 3 Divide each by smaller. If ratio not close to whole then double mole Ratio 2 parts Fe to 3 part O so Fe,O 70% Fe 70g Fe / 55.8 g/mol = 1.25/1.25 = 1 moles x 2 = 2 1. 75% carbon, 25% hydrogen Percent 25% H 75% C grams divide by molar mass Moles 25 g H / 75g C/ 1 g/mol 12g/mol = 25 moles/6.25 = 4 = 6.25 moles/6.25 = 1 Divide each by smaller mole Ratio 4 parts H to 1 part C so CH, 2. 52.7% potassium, 47.3% chlorine 3. 22.1% aluminum, 25.4% phosphorus, 52.5% oxygen 4. 13% magnesium, 87% bromine 5. 32.4% sodium, 22.5 % Sulfur, 45.1 % oxygen. 6. 25.3% copper, 12.9% sulfur, 25.7% oxygen, 36.1 % water 7. 22.3 % oxygen, 77.7 % iron
Transcribed Image Text:Chem Activity 055 Determining Empirical Formulas What is the empirical formula (lowest whole number ratio) of the compounds below? Example 1: 11% H, 88% o Percent grams divide by molar mass = Moles Divide each by smaller 1 g/mol 16g/mol *Assume that 100 % is 100 grams = 11 moles/5.5 = 2 = 5.5 moles/5.5 = 1 mole Ratio *11 g H / 88% O = *88 g O / 11% H = 2 parts H to 1 part O so Example 2: 30% O, 70% Fe Percent grams divide by molar mass = Moles 30% O, so 30 g0 / 16 g/mol = 1.875/1.25 = 1.5 moles x 2 = 3 Divide each by smaller. If ratio not close to whole then double mole Ratio 2 parts Fe to 3 part O so Fe,O 70% Fe 70g Fe / 55.8 g/mol = 1.25/1.25 = 1 moles x 2 = 2 1. 75% carbon, 25% hydrogen Percent 25% H 75% C grams divide by molar mass Moles 25 g H / 75g C/ 1 g/mol 12g/mol = 25 moles/6.25 = 4 = 6.25 moles/6.25 = 1 Divide each by smaller mole Ratio 4 parts H to 1 part C so CH, 2. 52.7% potassium, 47.3% chlorine 3. 22.1% aluminum, 25.4% phosphorus, 52.5% oxygen 4. 13% magnesium, 87% bromine 5. 32.4% sodium, 22.5 % Sulfur, 45.1 % oxygen. 6. 25.3% copper, 12.9% sulfur, 25.7% oxygen, 36.1 % water 7. 22.3 % oxygen, 77.7 % iron
8. 46.3 % Sulfur, 53.7 % iron 9. 29 % oxygen, 8.5 % nitrogen, 62.5% lead 10. 13.7 % sulfur, 27.4 % oxygen, 58.8 % barim e 56 on next page em Activity 056 Determining Molecular Formulas (" e the problems below The empirical formula of a compound is NO SA
Transcribed Image Text:8. 46.3 % Sulfur, 53.7 % iron 9. 29 % oxygen, 8.5 % nitrogen, 62.5% lead 10. 13.7 % sulfur, 27.4 % oxygen, 58.8 % barim e 56 on next page em Activity 056 Determining Molecular Formulas (" e the problems below The empirical formula of a compound is NO SA
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