Stoichiometry Chapter 3 E-mail: benzene4president@gmail.com Web-site: http://clas.sa.ucsb.edu/staff/terri/

Stoichiometry - Ch. 3 1. For a new element, 67.16% is an isotope with mass 280.8 amu, 2.76% is an isotope with mass 283.7 amu and 30.08% is an isotope with mass 284.8 amu. Calculate the average atomic mass for this new element?

Stoichiometry - Ch. 3 Average Atomic Mass = (fraction of isotope A)(mass of isotope A) + (fraction of isotope B)(mass of isotope B) + etc.

Stoichiometry - Ch. 3 2. For which of the following compounds does 1.00 g represent 3.32 × 10-2 mol? a. NO2 b. H2O c. C2H6 d. NH3 e. CO

Stoichiometry - Ch. 3 1 mole of PCl5 208.239 g of PCl5 5(35.453) g of Cl 30.974 g of P 1 mole of PCl5 5 moles of Cl P 6.022x1023 molecules 6(6.022x1023) total atoms P atoms 5(6.022x1023)Cl atoms

Stoichiometry - Ch. 3 3. If a sample of diatomic element weighs 131.3 g and contains 4.162x1024 atoms. Identify the element.

Stoichiometry - Ch. 3 4. If you have 0.63 mg of H2SO4 a. How many H2SO4 molecules are in your sample? b. How many oxygen atoms?

Stoichiometry - Ch. 3 5. An alkali metal oxide contains 83.01% metal by mass. Determine the identity of the metal. How many grams of oxygen are in a 25.0 g sample of the metal oxide?

Percent by mass = (mass of 1 part)x100 (Total mass) Stoichiometry - Ch. 3 Percent by mass = (mass of 1 part)x100 (Total mass)

Stoichiometry - Ch. 3 6. Compound X2Y is 60% X by mass. Calculate the percent Y by mass of the compound XY3?

Stoichiometry - Ch. 3 7. Tryptophan is an amino acid well known for its sleep inducing properties. Tryptophan is 64.7% carbon, 5.9% hydrogen, 13.7% nitrogen and 15.7% oxygen. What is the empirical formula for tryptophan?

Stoichiometry - Ch. 3 Empirical Formula ⇒ The lowest whole number molar ratio of the elements in a compound 1. Convert given values into moles 2. Divide all moles by the smallest mole value 3. If you have all whole numbers you have the EF – if not try multiplying them all by 2 or 3 etc.

Stoichiometry - Ch. 3 8. The empirical formula for xylene is C4H5 and xylene has a molar mass of 106.16 g/mol. Determine the molecular formula for xylene.

Stoichiometry - Ch. 3 Molecular Formula ⇒ The actual molar ratio of the elements in a compound – it is some multiple of the empirical formula (x1, x2 etc) 1. Derive empirical formula 2. Determine the empirical mass 3. (Molar mass)/(empirical mass) = multiple 4. Multiply the empirical formula by the multiple

Stoichiometry - Ch. 3 9. A 0.4647-g sample of a compound known to contain only carbon, hydrogen, and oxygen was burned in oxygen to yield 0.8635 g of CO2 and 0.1767 g of H2O. What is the empirical formula of the compound?

Stoichiometry - Ch. 3 10. A compound contains only carbon, hydrogen, nitrogen, and oxygen. Combustion of 0.157 g of the compound produced 0.213 g of CO2 and 0.0310 g of H2O. In another experiment, 0.103 g of the compound produced 0.0230 g of NH3 (assume all of the N ends up in the ammonia). What is the empirical formula for the compound?

Stoichiometry - Ch. 3 11. Consider the following unbalanced reaction: NH3 + O2  NO2 + H2O a. How many moles of oxygen gas are required to make 12.8 moles of nitrogen dioxide? b. How many grams of water can be produced from 9.64 g of ammonia? c. Identify the limiting reagent if 3 moles of ammonia is combined with 5 moles of oxygen d. Identify the limiting reagent if 10 g of ammonia is combined with 28 g of oxygen

Methodology for Reaction Stoichiometry Problems Stoichiometry - Ch. 3 Methodology for Reaction Stoichiometry Problems 1. Write a balanced chemical reaction 2. Convert given value(s) into moles (you may have to ID the limiting reagent – next slide) 3. Use reaction coefficients as a molar ratio 4. Convert moles of your unknown into the desired units

Stoichiometry - Ch. 3 Limiting Reagent ⇒ Limits the amount of product that is produced due to running out 1st The limiting reagent is used to determine the maximum yield of product/s aka the theoretical yield and the maximum consumption of reactants Identifying Limiting Reagents: 1. Convert all given values into moles 2. Divide each mole value by the coefficient 3. The smallest number identifies the LR

Ca3(PO4)2 + SiO2 + C  CaSiO3 + P4 + CO Stoichiometry - Ch. 3 12. Phosphorus can be prepared from calcium phosphate by the following unbalanced reaction: Ca3(PO4)2 + SiO2 + C  CaSiO3 + P4 + CO a. How many grams of carbon monoxide can be produced from a mixture of 10g of each reactant? b. What is the percent yield if 0.282 g of CO were obtained? c. How many grams of excess reactant remains?

Percent yield = (Actual yield)x(100) (Theoretical yield) Stoichiometry - Ch. 3 Percent yield = (Actual yield)x(100) (Theoretical yield)

Stoichiometry - Ch. 3 13. How many grams of fluorine are required if you want to produce 83 g of PF3 if the reaction has 63.2% yield? P4 + F2  PF3 (unbalanced)

Stoichiometry - Ch. 3 You have completed ch. 3

Ch. 3 – Answer Key 1. For a new element, 67.16% is an isotope with mass 280.8 amu, 2.76% is an isotope with mass 283.7 amu and 30.08% is an isotope with mass 284.8 amu. Calculate the average atomic mass for this new element? Average atomic mass = (0.6716)(280.8 amu) + (0.0276)(283.7 amu) + (0.3008)(284.8 amu) = 282.1 amu 2. For which of the following compounds does 1.00 g represent 3.32 × 10-2 mol? Molar mass is a useful value for identification Molar mass = (1.00g)/ (3.32 × 10-2 mol) = 30.1g/mol => C2H6

Ch. 3 – Answer Key 3. If a sample of diatomic element weighs 131.3 g and contains 4.162x1024 atoms. Identify the element. Diatomic tells us the formula for the element is X2. To get the molar mass you need the grams (given) and the moles (not given). 4.162x1024 atoms 1molecule X2 1mole X2 = 3.456 mole X2 Molar mass = 131.3g/3.456 mole = 38.0g/mol => F2 2 atoms 6.022x1023 molecules

Ch. 3 – Answer Key 4. If you have 0.63 mg of H2SO4 a. How many H2SO4 molecules are in your sample? The molar mass of H2SO4 = 2(1.01) + 32.06 + 4(16.0) = 98.08 g/mol 0.63 mg 1.0 g 1 mole 6.022x1023 molecules = 3.9x1018 molecules b. How many oxygen atoms are in your sample? 3.9x1018 molecules 4 oxygen atoms = 1.5x10 O atoms 1000 mg 98.08 g H2SO4 1 mole 1 H2SO4 molecule

Ch. 3 – Answer Key 5. An alkali metal oxide contains 83.01% metal by mass. Determine the identity of the metal. How many grams of oxygen are in a 25.0 g sample of the metal oxide? Since an alkali metal has a charge of 1+ the chemical formula for the metal oxide is M2O If the molar mass of the metal is denoted by x ⇒ 83.01 = ((2x)/(2x + 16))(100) x = 39.1g/mol => K Potassium is the unknown metal

Ch. 3 – Answer Key 6. Compound X2Y is 60% X by mass. Calculate the percent Y by mass of the compound XY3? If you have 100 g of X2Y there would be 60g of X and 40g of Y For XY3 since it has only one X atom you can think of that as ½(60g) or 30 g of X and since there’s three Y atoms you can think of that as 3(40g) or 120g of Y. So in XY3 there’s 30g of X for every 120g of Y – so %Y = (120g/150g)100 = 80%

Ch. 3 – Answer Key 7. Tryptophan is an amino acid well known for its sleep inducing properties. Tryptophan is 64.7% carbon, 5.9% hydrogen, 13.7% nitrogen and 15.7% oxygen. What is the empirical formula for tryptophan? First we need to get a moles for the molar ratio If we have 100 g sample of typtophan ⇒ 64.7 g C/12.01 g/mol = 5.347 mol C 5.9 g H/1.008 g/mol = 5.853 mol H 13.7 g N/14.01 g/mol = 0.978 mol N 15.7 g O/16 g/mol = 0.981 mol O continue to next slide...

Ch. 3 – Answer Key 7. …continued Divide each by the smallest mole value to simplify the ratio ⇒ 5.347 mol C/0.978 = 5.5 mol C 5.853 mol H/0.978 = 6 mol H 0.978 mol N/0.978 = 1 mol N 0.981 mol O/0.978 = 1 mol O Double each to get whole numbers ⇒ 11 mol C:12 mol H :2 mol N:2 mol O Empirical Formula ⇒ C11H12N2O2

Ch. 3 – Answer Key 8. The empirical formula for xylene is C4H5 and xylene has a molar mass of 106.16 g/mol. Determine the molecular formula for xylene. First determine the molar mass of the empirical formula ⇒ 4(12.01)g/mol + 5(1.008)g/mol = 53.08 g/mol Divide molar mass of empirical formula into molar mass of the compound ⇒ (106.16g/mol)/(53.08g/mol) = 2 Multiply the empirical formula by 2⇒ Molecular formula ⇒ C8H10

Ch. 3 – Answer Key CxHyOz + O2  CO2 + H2O ? 9. A 0.4647-g sample of a compound known to contain only carbon, hydrogen, and oxygen was burned in oxygen to yield 0.8635 g of CO2 and 0.1767 g of H2O. What is the empirical formula of the compound? CxHyOz + O2  CO2 + H2O All of the carbon in the compound will end up in the CO2 and all of the hydrogen will end up in the water the oxygen is unpredictable so we need determine how much C and H there is in our compound ? …continue to next slide

Ch. 3 – Answer Key 9. …continued C ⇒ 0.8635 g CO2 12.01 g C H ⇒ 0.1767 g H2O O ⇒ 0.4647 g cmpd – 0.2356 g C – 0.01977 g H = 0.2093 g O …continue to next slide 12.01 g C = 0.2356 g C 44.01 g CO2 2.016 g H = 0.01977 g H 18.016 g H2O

Ch. 3 – Answer Key 9. …continued Covert each to moles and divide by smallest value ⇒ (0.2356 g C)/(12.01 g/mol) = 0.01962 mol C/0.01308 = 1.5 mol C (0.01977 g H)/(1.008 g/mol) = 0.01961 mol H/0.01308 = 1.5 mol H (0.2093 g O)/(16 g/mol) = 0.01308 mol O/0.01308 = 1 mol O Double each value 3 mol C:3 mol H:2 mol O Empirical formula ⇒ C3H3O2

Ch. 3 – Answer Key CaHbNcOd + O2  CO2 + H2O + ? ? 10. A compound contains only carbon, hydrogen, nitrogen, and oxygen. Combustion of 0.157 g of the compound produced 0.213 g of CO2 and 0.0310 g of H2O. In another experiment, 0.103 g of the compound produced 0.0230 g of NH3. What is the empirical formula for the compound? CaHbNcOd + O2  CO2 + H2O + ? In the combustion reaction you can map out that all of the C goes into the CO2 and all of the H goes into the water but we are unable to map the path of the O – in the second expt we can only map out the N …continue to next slide ? CaHbNcOd + ?  NH3 + ?

Ch. 3 – Answer Key 10. …continued C ⇒ 0.213 g of CO2 12.01 g C H ⇒ 0.0310 g of H2O N ⇒ 0.0230 g of NH3 12.01 g C = 0.05813 g C (1st experiment) 44.01 g CO2 2.016 g H = 0.003469 g H (1st experiment) 18.016 g H2O 14.01 g N = 0.01892 g N (2nd experiment) 17.034 g NH3 Since the nitrogen was determined from a different experiment we can use % by mass to figure out the mass of nitrogen in the 1st experiment …continue to next slide

Ch. 3 – Answer Key 10. …continued % N = ((0.01892 g N)/(0.103 g cmpd)) x 100 = 18.37 % N So in the 1st sample ⇒ (0.1837)x(0.157g cmpd) = 0.02883 g N Oxygen is the remainder ⇒ 0.157 g cmpd – 0.01583 g C – 0.003469 g H – 0.02884 g N =0.06656g O Covert each to moles and divide by the smallest value (0.05813 g C)/(12.01 g/mol) = 0.00484 mol C/0.002058 = 2.35 mol C (0.003469 g H)/(1.008 g/mol) = 0.003441 mol H/0.002058 = 1.67mol H (0.02883 g N)/(14.01 g/mol) = 0.002058 mol N/0.002058 = 1mol N (0.06656 g O)/(16 g/mol) = 0.00416 mol O/0.002058 = 2 mol O multiply by 3 to get the empirical formula⇒ C7H5N3O6

Ch. 3 – Answer Key 11. Consider the following unbalanced reaction: 4 NH3 + 7 O2  4 NO2 + 6 H2O a. How many moles of oxygen gas are required to make 12.8 moles of nitrogen dioxide? 12.8 mol NO2 7 mol O2 = 22.4 mol O2 b. How many grams of water can be produced from 9.64 g of ammonia? 9.64 g NH3 1mol NH3 6 mol H2O 18.02 g H2O = 15.3 g H2O 4 mol NO2 17.04 g NH3 4 mol NH3 1 mol H2O Continue to next slide…

Ch. 3 – Answer Key convert to moles then divide by molar coefficient 11. …continued c. Identify the limiting reagent if 3 moles of ammonia is combined with 5 moles of oxygen divide each mole by the molar coefficient and look for smaller value (3 mol NH3)/4 = 0.75 vs. (5 mol O2)/7 = 0.71 ⇒ O2 is the LR d. Identify the limiting reagent if 10. g of ammonia is combined with 28 g of oxygen convert to moles then divide by molar coefficient (10 g NH3)/(17.034 g/mol) = 0.587 mol NH3/4 = 0.147 vs. (28 g O2)/(32 g/mol) = 0.875 mol O2/7 = 0.125 ⇒ O2 is the LR

Ca3(PO4)2 + SiO2 + C  CaSiO3 + P4 + CO Ch. 3 – Answer Key 12. Phosphorus can be prepared from calcium phosphate by the following unbalanced reaction: Ca3(PO4)2 + SiO2 + C  CaSiO3 + P4 + CO a. How many grams of carbon monoxide can be produced from a mixture of 10g of each reactant? First ⇒ balance the reaction 2 Ca3(PO4)2 + 6 SiO2 + 10 C  6 CaSiO3 + P4 + 10 CO Next ⇒ determine the limiting reagent (10 g Ca3(PO4)2)/(310.18 g/mol) = 0.0322 mol Ca3(PO4)2/2 = 0.0161 (10 g SiO2)/(60.09 g/mol) = 0.166 mol SiO2/6 = 0.0277 (10 g C)/(12.01g/mol) = 0.833 mol C/10 = 0.0833 Ca3(PO4)2 is the LR 0.0322 mol Ca3(PO4)2 10 mol CO 28.01 g CO = 4.51 g CO (theoretical yield) 2 mol Ca3(PO4)2 1 mol CO

Ch. 3 – Answer Key 12. …continued b. What is the percent yield if 0.282 g of CO were obtained? %yield = (0.282g/4.51g)100 = 6.25% c. How many grams of excess reactant remains? The LR determines how much is consumed for each of the other reactants 0.0322 mol Ca3(PO4)2 6 mol SiO2 60.09 g SiO2 = 5.80 g SiO2 is consumed 0.0322 mol Ca3(PO4)2 10 mol C 12.01 g C = 1.9 g C is consumed 2 mol Ca3(PO4)2 1 mol SiO2 10g – 5.8g = 4.2g SiO2 are leftover 2 mol Ca3(PO4)2 1 mol C 10g – 1.9g = 8.1g C are leftover

Ch. 3 – Answer Key 13. How many grams of fluorine are needed to produce 83 g of phosphorus trifluoride, if the reaction has 63.2% yield? P4 + F2  PF3 (unbalanced) First ⇒ balance reaction ⇒ P4 + 6 F2  4 PF3 Determine theoretical yield from the actual and the % yield ⇒ 63.2% = (83 g/Theo)x100 ⇒ Theoretical yield = 131.3 g PF3 Use theoretical yield to determine grams of F2 131.3 g PF3 1 mol PF3 6 mol F2 38 g F2 = 85.1 g F2 87.97 g PF3 4 mol PF3 1 mol F2