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3. Mg(NO,), Mass of Mg- 24.3 x 1- 243 g/ Mass of N- Mass of O-

3. Mg(NO,), Mass of Mg- 24.3 x 1- 243 g/ Mass of N- Mass of O-

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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sClas Assignment 301 Perce Com Chen Activiy ose Pret mpo Moles and Empirical Fommulas CwwLA MY. yeof Example 2: Se if you had 100 g of water then Haound by 11% which is 11x 100 Os tound by 88% which is Mx 100 88 grams O Example 1: ,O Mass of H 1.0x2-20g Mass of O 16.0x 116.0 Total HO So 2.0 gr18 gH,O 11%H 16 gort8 g H0 88% O 0 88% 1% mssing due to rounding error grams H H 11% L KMaO 39.1 x 1- 39.1g/157.1g-25% Mass of Mn- 54.9 x 1- 54.9g/157.1g-35% 16.0 x 4-64.0g/157.1g 41% 157.1g Mass of K- Mass of O- 2. HCI Mass of H-1.0 x 1- 1.0g/364g 2.7% Mass of Cl- 35.4 x 1- 35.4g/36 4g- 97% 36,4 3. Mg(NO,), Mass of Mg- 24.3 x 1- 24.3 g/ Mass of N- Mass of O- 4 (NH,)PO, N- P- 5. AL(SO) Al S- 6 How many grams osygen can be produced from the decomposition of 100 G of KCIO 7. How much iron can be recovered from 25.0 g of Fe0?
Transcribed Image Text:sClas Assignment 301 Perce Com Chen Activiy ose Pret mpo Moles and Empirical Fommulas CwwLA MY. yeof Example 2: Se if you had 100 g of water then Haound by 11% which is 11x 100 Os tound by 88% which is Mx 100 88 grams O Example 1: ,O Mass of H 1.0x2-20g Mass of O 16.0x 116.0 Total HO So 2.0 gr18 gH,O 11%H 16 gort8 g H0 88% O 0 88% 1% mssing due to rounding error grams H H 11% L KMaO 39.1 x 1- 39.1g/157.1g-25% Mass of Mn- 54.9 x 1- 54.9g/157.1g-35% 16.0 x 4-64.0g/157.1g 41% 157.1g Mass of K- Mass of O- 2. HCI Mass of H-1.0 x 1- 1.0g/364g 2.7% Mass of Cl- 35.4 x 1- 35.4g/36 4g- 97% 36,4 3. Mg(NO,), Mass of Mg- 24.3 x 1- 24.3 g/ Mass of N- Mass of O- 4 (NH,)PO, N- P- 5. AL(SO) Al S- 6 How many grams osygen can be produced from the decomposition of 100 G of KCIO 7. How much iron can be recovered from 25.0 g of Fe0?
10. 13.7 % sulfur, 27.4 % oxygen, 58.8 % barim See 56 on next page Chem Activity 056 Determining Molecular Formulas (True Formulas) Solve the problems below 1. The empirical formula of a compound is NO,. Its molecular mass is 92 g/mol. What is its molecular formula? Empirical formula is NO,, and the molar mass of NO, is 46 g/mole. So 92 g/mol/46 g/mol is 2. So 2 x NO,NO. which is the molecular formula 2. The empirical formula of a compound in CH,. Its molecular mass in 70 g/mol. What is the molecular formula? 3. A compound is found to be 40.0% carbon, 6.7% hydrogen and 53.55 % oxygen. Its molecular mass is 60. g/mol What is the molecular formula? 4. A compound is 64.9% carbon, 13.5% hydrogen and 21.6% oxygen. Its molecular mass is 74 g/mol. What is the molecular formula? 5. A compound is 54.5% carbon, 9.1% hydrogen and 36.4 % oxygen. Its molecular mass is 88 g/mol. What is the molecular formula?
Transcribed Image Text:10. 13.7 % sulfur, 27.4 % oxygen, 58.8 % barim See 56 on next page Chem Activity 056 Determining Molecular Formulas (True Formulas) Solve the problems below 1. The empirical formula of a compound is NO,. Its molecular mass is 92 g/mol. What is its molecular formula? Empirical formula is NO,, and the molar mass of NO, is 46 g/mole. So 92 g/mol/46 g/mol is 2. So 2 x NO,NO. which is the molecular formula 2. The empirical formula of a compound in CH,. Its molecular mass in 70 g/mol. What is the molecular formula? 3. A compound is found to be 40.0% carbon, 6.7% hydrogen and 53.55 % oxygen. Its molecular mass is 60. g/mol What is the molecular formula? 4. A compound is 64.9% carbon, 13.5% hydrogen and 21.6% oxygen. Its molecular mass is 74 g/mol. What is the molecular formula? 5. A compound is 54.5% carbon, 9.1% hydrogen and 36.4 % oxygen. Its molecular mass is 88 g/mol. What is the molecular formula?
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