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3. Mg(NO;), Mass of Mg = 24.3 x 1= 24.3 g/ %3D Mass of N = %3D Mass of O = 4. (NH,),PO4 %3D 5. Al,(SO,), Al = S = NHPO

3. Mg(NO;), Mass of Mg = 24.3 x 1= 24.3 g/ %3D Mass of N = %3D Mass of O = 4. (NH,),PO4 %3D 5. Al,(SO,), Al = S = NHPO

Introductory Chemistry: A Foundation
Introductory Chemistry: A Foundation
9th Edition
ISBN:9781337399425
Author:Steven S. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Donald J. DeCoste
Chapter8: Chemical Composition
Section: Chapter Questions
Problem 21ALQ: True or false? The atom with the largest subscript in a formula is the atom with the largest percent...
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in Class Assignment 301 Percent Composition, Moles and Empirical Formulas Chem Activity 054 Percentage Composition Namefenza ryinDate: Determine the percentage composition (By Mass) of the compounds below: Class: 2A Teacher: MY. Lppin Example 2: So if you had 100 g of water then: H is found by 11% which is.11 x 100 = 11 grams H O is found by 88% which is .88x 100 = 88 grams O Example 1: H,0 Mass of H = 1.0 x 2 = 2.0 g Mass of O = 16.0 x 1= 16 g Total H,O So 2.0 gH/18 gH,O = 11 % H 16 gO/18 g H2O = 88 % O 1 % missing due to rounding error. = 18 g H = 11 % O = 88 % 1. KMNO, Mass of K = 39.1 x 1= 39.1g/157.1g= 25% Mass of Mn = 54.9 x 1= 54.9g/157.1g 35% Mass of O = 16.0 x 4= 64.0g/157.1g= 41% 157.1g 2. HCI Mass of H = 1.0 x 1= 1.0g/36.4g = 2.7% Mass of Cl = 35.4 x 1= 35.4g/36.4g = 97% 36.4 3. Mg(NO3), Mass of Mg = 24.3 x 1= 24.3 g/ Mass of N = Mass of O = 4. (NH;),PO4 N = H = P = 5. Al,(SO,), Al = S = How many gran
Transcribed Image Text:in Class Assignment 301 Percent Composition, Moles and Empirical Formulas Chem Activity 054 Percentage Composition Namefenza ryinDate: Determine the percentage composition (By Mass) of the compounds below: Class: 2A Teacher: MY. Lppin Example 2: So if you had 100 g of water then: H is found by 11% which is.11 x 100 = 11 grams H O is found by 88% which is .88x 100 = 88 grams O Example 1: H,0 Mass of H = 1.0 x 2 = 2.0 g Mass of O = 16.0 x 1= 16 g Total H,O So 2.0 gH/18 gH,O = 11 % H 16 gO/18 g H2O = 88 % O 1 % missing due to rounding error. = 18 g H = 11 % O = 88 % 1. KMNO, Mass of K = 39.1 x 1= 39.1g/157.1g= 25% Mass of Mn = 54.9 x 1= 54.9g/157.1g 35% Mass of O = 16.0 x 4= 64.0g/157.1g= 41% 157.1g 2. HCI Mass of H = 1.0 x 1= 1.0g/36.4g = 2.7% Mass of Cl = 35.4 x 1= 35.4g/36.4g = 97% 36.4 3. Mg(NO3), Mass of Mg = 24.3 x 1= 24.3 g/ Mass of N = Mass of O = 4. (NH;),PO4 N = H = P = 5. Al,(SO,), Al = S = How many gran
8. How much silver can be produced from 125 g of Ag,S? 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 = H %LL 88% O = *88 g O / 1 g/mol 16g/mol = 11 moles/5.5 = 2 = 5.5 moles/5.5 = 1 mole Ratio 2 parts H to 1 part O so *11 g H I *Assume that 100 % is 100 grams Example 2: 30% O, 70% Fe Percent grams divide by molar mass = Moles 30% O, so 30 g0 16 g/mol 70% Fe Divide each by smaller. If ratio not close to whole then double mole Ratio = 1.875/1.25 = 1,5 moles x 2 = 3 70g Fe / 55.8 g/mol = 1.25/1.25 = 1 moles x 2 = 2 2 parts Fe to 3 part O so 1. 75% carbon, 25% hydrogen Percent grams divide by molar mass = Moles Divide each by smaller H %S 25 g H / 75g C/ 1 g/mol 12g/mol = 25 moles/6.25 = 4 = 6.25 moles/6.25 = 1 mole Ratio 4 parts H to 1 part C so CH 75% C 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:8. How much silver can be produced from 125 g of Ag,S? 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 = H %LL 88% O = *88 g O / 1 g/mol 16g/mol = 11 moles/5.5 = 2 = 5.5 moles/5.5 = 1 mole Ratio 2 parts H to 1 part O so *11 g H I *Assume that 100 % is 100 grams Example 2: 30% O, 70% Fe Percent grams divide by molar mass = Moles 30% O, so 30 g0 16 g/mol 70% Fe Divide each by smaller. If ratio not close to whole then double mole Ratio = 1.875/1.25 = 1,5 moles x 2 = 3 70g Fe / 55.8 g/mol = 1.25/1.25 = 1 moles x 2 = 2 2 parts Fe to 3 part O so 1. 75% carbon, 25% hydrogen Percent grams divide by molar mass = Moles Divide each by smaller H %S 25 g H / 75g C/ 1 g/mol 12g/mol = 25 moles/6.25 = 4 = 6.25 moles/6.25 = 1 mole Ratio 4 parts H to 1 part C so CH 75% C 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
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