# Stoichiometry Chapter 3. Atomic Mass 1961: Atomic Mass is based on 12 C 1961: Atomic Mass is based on 12 C 12 C is assigned a mass of EXACTLY 12 AMU 12.

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Stoichiometry Chapter 3

Atomic Mass 1961: Atomic Mass is based on 12 C 1961: Atomic Mass is based on 12 C 12 C is assigned a mass of EXACTLY 12 AMU 12 C is assigned a mass of EXACTLY 12 AMU Most Accurate Way to compare masses is to use a mass spectrometer Most Accurate Way to compare masses is to use a mass spectrometer Results in ratio of masses like 13 C/ 12 C Results in ratio of masses like 13 C/ 12 C Apply ratio to mass of 12 C to get atomic mass. Apply ratio to mass of 12 C to get atomic mass. Atomic Mass on Periodic Table is a weighted average of the isotopes Atomic Mass on Periodic Table is a weighted average of the isotopes 98.89% 12 C, 1.11% 13 C, negligible amounts of 14 C 98.89% 12 C, 1.11% 13 C, negligible amounts of 14 C (.9889)(12) + (.011)(13.0034) = 12.01amu (.9889)(12) + (.011)(13.0034) = 12.01amu

The Mole Mole: the number equal to the number of carbon atoms in exactly 12 grams of pure 12 C Mole: the number equal to the number of carbon atoms in exactly 12 grams of pure 12 C Avogadro’s number: 6.022x10 23 units = 1 mole Avogadro’s number: 6.022x10 23 units = 1 mole 1 mole of an element weighs exactly the atomic mass of the element in grams 1 mole of an element weighs exactly the atomic mass of the element in grams 1 mole of a molecule weighs exactly the atomic mass of the molecule in grams 1 mole of a molecule weighs exactly the atomic mass of the molecule in grams

Calculating Number of Atoms A silicon chip for a computer has a mass of 5.68 mg. How many atoms of silicon are in the chip? A silicon chip for a computer has a mass of 5.68 mg. How many atoms of silicon are in the chip? 1.Convert milligrams to grams 2.Convert grams to moles 3.Convert moles to atoms

Percent Composition of Compounds 2 common ways to describe the composition of a compound: 2 common ways to describe the composition of a compound: Number of constituent atoms Number of constituent atoms Percent composition by mass Percent composition by mass Example: % composition of ethanol (C 2 H 5 OH) Example: % composition of ethanol (C 2 H 5 OH) 2 mol C x 12.01 g/mol = 24.02g C 2 mol C x 12.01 g/mol = 24.02g C 6 mol H x 1.008 g/mol = 6.048 g H 6 mol H x 1.008 g/mol = 6.048 g H 1 mol O x 16.00 g/mol – 16.00 g O 1 mol O x 16.00 g/mol – 16.00 g O % C = 24.02g/(24.02g+6.048g+16.00g) x 100 = 52.14% C % C = 24.02g/(24.02g+6.048g+16.00g) x 100 = 52.14% C %H = 6.048g/46.07g x 100 = 13.13% H %H = 6.048g/46.07g x 100 = 13.13% H %O = 16.00g/46.07g x 100 = 34.73% O %O = 16.00g/46.07g x 100 = 34.73% O

Determining Formula of Compounds You have 0.1156g of a new compound composed of C, H, and N You have 0.1156g of a new compound composed of C, H, and N You decomposed or reacted it with water to get CO 2, and H 2 O, which was collected and weighed. You decomposed or reacted it with water to get CO 2, and H 2 O, which was collected and weighed. 0.1638g CO 2 0.1638g CO 2 0.1676g H 2 O 0.1676g H 2 O How can you calculate the formula of the compound? How can you calculate the formula of the compound? How much carbon was in the original compound? How much carbon was in the original compound? How much nitrogen was in the original compound? How much nitrogen was in the original compound?

Determining Formula of Compounds Use % by mass of the product compounds: Use % by mass of the product compounds: C: 1 mol x 12.01 g/mol = 12.01g C C: 1 mol x 12.01 g/mol = 12.01g C O: 2 mole x16.00g/mol = 32.00g O O: 2 mole x16.00g/mol = 32.00g O Molar mass CO2 = 44.01 g/mol Molar mass CO2 = 44.01 g/mol 0.1638 g CO2 x 12.01gC/44.01gCO2 = 0.004470 g C 0.1638 g CO2 x 12.01gC/44.01gCO2 = 0.004470 g C 0.00470g CO2/0.1156g compound x 100% = 38.67% C 0.00470g CO2/0.1156g compound x 100% = 38.67% C 0.1676 g H2O x 2.016 g H/18.02 g H2O = 0.001875 g H 0.1676 g H2O x 2.016 g H/18.02 g H2O = 0.001875 g H 0.001875 g H /0.1156g compound x 100% = 16.22% H 0.001875 g H /0.1156g compound x 100% = 16.22% H Rest (100 – 38.67 – 16.22) is N (45.11%) Rest (100 – 38.67 – 16.22) is N (45.11%)

Calculate Empirical Formula From % by mass, calculate Empirical Formula From % by mass, calculate Empirical Formula ASSUME YOU HAVE 100 g OF COMPOUND ASSUME YOU HAVE 100 g OF COMPOUND 38.67 g C x 1 mol/12.01 g = 3.220 mol C 38.67 g C x 1 mol/12.01 g = 3.220 mol C 16.22 g H x 1 mol/1.008 g = 16.09 mol H 16.22 g H x 1 mol/1.008 g = 16.09 mol H 45.11 g N x 1 mol/14.01 g = 3.219 mol N 45.11 g N x 1 mol/14.01 g = 3.219 mol N Find smallest WHOLE-NUMBER ratios of elements: Find smallest WHOLE-NUMBER ratios of elements: 3.220 mol C/3.219 mol N = 1.000 = 1:1 3.220 mol C/3.219 mol N = 1.000 = 1:1 16.09 mol O/3.220 mol C = 4.997 = 5:1 16.09 mol O/3.220 mol C = 4.997 = 5:1 Formula would be: CH 5 N Formula would be: CH 5 N Or any whole number multiple of those elements Or any whole number multiple of those elements

Actual Molecular Formula Must determine Actual molecular mass experimentally Must determine Actual molecular mass experimentally Suppose compound is known to have mass of 62.12 g/mol Suppose compound is known to have mass of 62.12 g/mol Actual mass/Empirical mass gives ratio Actual mass/Empirical mass gives ratio 62.12g/mol / 31.06 G/mol = 2.0 62.12g/mol / 31.06 G/mol = 2.0 Actual formula is (CH 5 N)x2 = C 2 H 10 N 2 Actual formula is (CH 5 N)x2 = C 2 H 10 N 2 Solve Questions 69-75 (one per group), page 119-120 in textbook. Solve Questions 69-75 (one per group), page 119-120 in textbook.

Limiting Reactants Some chemical processes are carried out with excess amounts of one or more reactants. Some chemical processes are carried out with excess amounts of one or more reactants. Chemically we learn about this in equilibrium Chemically we learn about this in equilibrium The excess reactants are usually the cheaper ones. The excess reactants are usually the cheaper ones. How do you handle questions where you have quantities of reactants that are not stoichiometrically balanced? How do you handle questions where you have quantities of reactants that are not stoichiometrically balanced? Concept of limiting reactant. Concept of limiting reactant.

Limiting Reactants How to solve the problem: How to solve the problem: Solve the problem once for each reactant that MIGHT be limiting. Solve the problem once for each reactant that MIGHT be limiting. Determine which solution has the lowest number of product moles Determine which solution has the lowest number of product moles That is your limiting reactant and maximum amount of product. That is your limiting reactant and maximum amount of product. Any other reactant above the stoichiometric amount of the limiting reactant is “excess” Any other reactant above the stoichiometric amount of the limiting reactant is “excess”

Limiting Reactant Solid lithium hydroxide is used in space vehicles to remove exhaled carbon dioxide from the living environment by forming solid lithim carbonate and liquid water. What mass of gaseous carbon dioxide can be absorbed by 1.00kg of lithium hydroxide? Solid lithium hydroxide is used in space vehicles to remove exhaled carbon dioxide from the living environment by forming solid lithim carbonate and liquid water. What mass of gaseous carbon dioxide can be absorbed by 1.00kg of lithium hydroxide? Step 1: Write the unbalanced equation: Step 1: Write the unbalanced equation: LiOH(s) + CO 2 (g) LiOH(s) + CO 2 (g)  Li 2 CO 3 (s) + H 2 O(l) Balance the equation: 2LiOH(s) + CO 2 (g) 2LiOH(s) + CO 2 (g)  Li 2 CO 3 (s) + H 2 O(l)

Limiting Reactant 2LiOH(s) + CO 2 (g) 2LiOH(s) + CO 2 (g)  Li 2 CO 3 (s) + H 2 O(l) Step 2: convert LiOH to moles Step 2: convert LiOH to moles Step 3: Determine amount of CO2 that reacts with a Given amount of LiOH Step 4: Calculate moles of CO2 that reacts with a given mass of LiOH

Limiting Reagent Step 5: Calculate mass of CO2 using molar mass Step 5: Calculate mass of CO2 using molar mass

Limiting Reactant Nitrogen gas can be prepared by passing gaseous ammonia over solid copper(II) oxide at high temperatures. The other products are solid copper and water vapor. If a sample of 18.1g of NH 3 is reacted with 90.4g of CuO, which is the limiting reactant? How manuy grams of N 2 will be formed? Nitrogen gas can be prepared by passing gaseous ammonia over solid copper(II) oxide at high temperatures. The other products are solid copper and water vapor. If a sample of 18.1g of NH 3 is reacted with 90.4g of CuO, which is the limiting reactant? How manuy grams of N 2 will be formed? Balanced equation: Balanced equation: 2NH 3 (g) + 3CuO(s)  N 2 (g) + 3 Cu(s) + 3H 2 O(g) 2NH 3 (g) + 3CuO(s)  N 2 (g) + 3 Cu(s) + 3H 2 O(g)

Limiting Reactant Compute moles of NH3 and CuO Compute moles of NH3 and CuO 18.1g NH 3 x 1mol NH 3 /17.03gNH 3 = 1.06mol NH 3 18.1g NH 3 x 1mol NH 3 /17.03gNH 3 = 1.06mol NH 3 90.4g CuO x 1 mol CuO/79.55g CuO = 1.14 mol CuO 90.4g CuO x 1 mol CuO/79.55g CuO = 1.14 mol CuO Use Mole ratio of CuO and NH3 to determine limiting reactant: Use Mole ratio of CuO and NH3 to determine limiting reactant: 1.06 mol NH 3 x 3mol CuO/2mol NH 3 = 1.59 mol CuO 1.06 mol NH 3 x 3mol CuO/2mol NH 3 = 1.59 mol CuO Because it takes 1.59 mole of CuO to react with 1.06 mole of NH 3, but we only have 1.14 mole of CuO, CuO is the limiting reactant. Because it takes 1.59 mole of CuO to react with 1.06 mole of NH 3, but we only have 1.14 mole of CuO, CuO is the limiting reactant. We will run out of CuO before we run out of NH 3 We will run out of CuO before we run out of NH 3 Verify: Verify: Balanced Equation: Mol CuO/Mol NH 3 = 3/2 = 1.5 Balanced Equation: Mol CuO/Mol NH 3 = 3/2 = 1.5 Amount Present: Mole CuO/Mol NH 3 = 1.14/1.06 = 1.08 Amount Present: Mole CuO/Mol NH 3 = 1.14/1.06 = 1.08

Limiting Reactant Calculate amount of N2 formed based on moles of CuO (limiting factor) present: Calculate amount of N2 formed based on moles of CuO (limiting factor) present: 1 mol N 2 /3 mol CuO x 1.14 mol CuO = 0.380 mol N 2 1 mol N 2 /3 mol CuO x 1.14 mol CuO = 0.380 mol N 2

% Yield Theoretical yield is the amount of product formed based on actual amounts and balanced equation stoichiometric calculations Theoretical yield is the amount of product formed based on actual amounts and balanced equation stoichiometric calculations Actual yield is the measured amount of product formed Actual yield is the measured amount of product formed % yield is actual yield/theoretical yield * 100 % yield is actual yield/theoretical yield * 100 Assignment: Questions 97a, 98c, 99-103 Assignment: Questions 97a, 98c, 99-103 One for each group One for each group

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