Download presentation

Presentation is loading. Please wait.

1
Stoichiometry Chapter 3 Web-site:

2
Stoichiometry - Ch For a new element, 67.16% is an isotope with mass amu, 2.76% is an isotope with mass amu and 30.08% is an isotope with mass amu. Calculate the average atomic mass for this new element?

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

4
Stoichiometry - Ch For which of the following compounds does 1.00 g represent 3.32 × mol? a. NO 2 b. H 2 O c. C 2 H 6 d. NH 3 e. CO

5
Stoichiometry - Ch g of PCl 5 5(35.453) g of Cl g of P g of P 1 mole of PCl 5 1 mole of PCl 5 5 moles of Cl 5 moles of Cl 1 mole of P 1 mole of P 6.022x10 23 molecules 6.022x10 23 molecules 6(6.022x10 23 ) total atoms 6.022x10 23 P atoms 6.022x10 23 P atoms 5(6.022x10 23 ) Cl atoms

6
Stoichiometry - Ch If a sample of diatomic element weighs g and contains 4.162x10 24 atoms. Identify the element.

7
Stoichiometry - Ch If you have 0.63 mg of H 2 SO 4 a. How many H 2 SO 4 molecules are in your sample? b. How many oxygen atoms?

8
Stoichiometry - Ch 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?

9
Stoichiometry - Ch. 3

10
6. Compound X 2 Y is 60% X by mass. Calculate the percent Y by mass of the compound XY 3 ?

11
Stoichiometry - Ch 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?

12
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. 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.

13
Stoichiometry - Ch The empirical formula for xylene is C 4 H 5 and xylene has a molar mass of g/mol. Determine the molecular formula for xylene.

14
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 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

15
Stoichiometry - Ch A g sample of a compound known to contain only carbon, hydrogen, and oxygen was burned in oxygen to yield g of CO 2 and g of H 2 O. What is the empirical formula of the compound?

16
Stoichiometry - Ch A compound contains only carbon, hydrogen, nitrogen, and oxygen. Combustion of g of the compound produced g of CO 2 and g of H 2 O. In another experiment, g of the compound produced g of NH 3 (assume all of the N ends up in the ammonia). What is the empirical formula for the compound?

17
Stoichiometry - Ch Consider the following unbalanced reaction: NH 3 + O 2 NO 2 + H 2 O 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

18
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 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

19
Stoichiometry - Ch. 3 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 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 Limiting Reagent ⇒ Limits the amount of product that is produced due to running out 1 st The limiting reagent is used to determine the maximum yield of product/s aka the theoretical yield and the maximum consumption of reactants Limiting Reagent ⇒ Limits the amount of product that is produced due to running out 1 st The limiting reagent is used to determine the maximum yield of product/s aka the theoretical yield and the maximum consumption of reactants

20
Stoichiometry - Ch Phosphorus can be prepared from calcium phosphate by the following unbalanced reaction: Ca 3 (PO 4 ) 2 + SiO 2 + C CaSiO 3 + P 4 + 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 g of CO were obtained? c. How many grams of excess reactant remains?

21
Stoichiometry - Ch. 3

22
13. How many grams of fluorine are required if you want to produce 83 g of PF 3 if the reaction has 63.2% yield? P 4 + F 2 PF 3 (unbalanced)

23
Stoichiometry - Ch. 3 You have completed ch. 3

24
1. For a new element, 67.16% is an isotope with mass amu, 2.76% is an isotope with mass amu and 30.08% is an isotope with mass 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) = amu 2. For which of the following compounds does 1.00 g represent 3.32 × mol? Molar mass is a useful value for identification Molar mass = (1.00g)/ (3.32 × mol) = 30.1g/mol => C 2 H 6 Ch. 3 – Answer Key

25
3. If a sample of diatomic element weighs g and contains 4.162x10 24 atoms. Identify the element. Diatomic tells us the formula for the element is X 2. To get the molar mass you need the grams (given) and the moles (not given) x10 24 atoms 1molecule X 2 1mole X 2 = mole X 2 Molar mass = 131.3g/3.456 mole = 38.0g/mol => F 2 2 atoms 6.022x10 23 molecules

26
Ch. 3 – Answer Key 4. If you have 0.63 mg of H 2 SO 4 a. How many H 2 SO 4 molecules are in your sample? The molar mass of H 2 SO 4 = 2(1.01) (16.0) = g/mol 0.63 mg 1.0 g 1 mole 6.022x10 23 molecules = 3.9x10 18 molecules b. How many oxygen atoms are in your sample? 3.9x10 18 molecules 4 oxygen atoms = 1.5x10 O atoms 1000 mg g H 2 SO 4 1 mole 1 H 2 SO 4 molecule

27
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 M 2 O If the molar mass of the metal is denoted by x ⇒ = ((2x)/(2x + 16))(100) x = 39.1g/mol => K Potassium is the unknown metal

28
Ch. 3 – Answer Key 6. Compound X 2 Y is 60% X by mass. Calculate the percent Y by mass of the compound XY 3 ? If you have 100 g of X 2 Y there would be 60g of X and 40g of Y For XY 3 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 XY 3 there’s 30g of X for every 120g of Y – so %Y = (120g/150g)100 = 80%

29
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 = mol C 5.9 g H/1.008 g/mol = mol H 13.7 g N/14.01 g/mol = mol N 15.7 g O/16 g/mol = mol O continue to next slide...

30
Ch. 3 – Answer Key 7. …continued Divide each by the smallest mole value to simplify the ratio ⇒ mol C/0.978 = 5.5 mol C mol H/0.978 = 6 mol H mol N/0.978 = 1 mol N 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 ⇒ C 11 H 12 N 2 O 2

31
Ch. 3 – Answer Key 8. The empirical formula for xylene is C 4 H 5 and xylene has a molar mass of 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 = 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 ⇒ C 8 H 10

32
Ch. 3 – Answer Key 9. A g sample of a compound known to contain only carbon, hydrogen, and oxygen was burned in oxygen to yield g of CO 2 and g of H 2 O. What is the empirical formula of the compound? C x H y O z + O 2 CO 2 + H 2 O ? All of the carbon in the compound will end up in the CO 2 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

33
Ch. 3 – Answer Key 9. …continued C ⇒ g CO 2 H ⇒ g H 2 O O ⇒ g cmpd – g C – g H = g O …continue to next slide g C g CO g H g H 2 O = g C = g H

34
Ch. 3 – Answer Key 9. …continued Covert each to moles and divide by smallest value ⇒ ( g C)/(12.01 g/mol) = mol C/ = 1.5 mol C ( g H)/(1.008 g/mol) = mol H/ = 1.5 mol H ( g O)/(16 g/mol) = mol O/ = 1 mol O Double each value 3 mol C:3 mol H:2 mol OEmpirical formula ⇒ C 3 H 3 O 2

35
Ch. 3 – Answer Key 10. A compound contains only carbon, hydrogen, nitrogen, and oxygen. Combustion of g of the compound produced g of CO 2 and g of H 2 O. In another experiment, g of the compound produced g of NH 3. What is the empirical formula for the compound? C a H b N c O d + O 2 CO 2 + H 2 O + ? ? C a H b N c O d + ? NH 3 + ? 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

36
Ch. 3 – Answer Key 10. …continued C ⇒ g of CO 2 H ⇒ g of H 2 O N ⇒ g of NH g C g CO 2 = g C (1 st experiment) g H g H 2 O = g H (1st experiment) g N g NH 3 = g N (2 nd experiment) Since the nitrogen was determined from a different experiment we can use % by mass to figure out the mass of nitrogen in the 1 st experiment …continue to next slide

37
10. …continued % N = (( g N)/(0.103 g cmpd)) x 100 = % N So in the 1 st sample ⇒ (0.1837)x(0.157g cmpd) = g N Oxygen is the remainder ⇒ g cmpd – g C – g H – g N = g O Covert each to moles and divide by the smallest value ( g C)/(12.01 g/mol) = mol C/ = 2.35 mol C ( g H)/(1.008 g/mol) = mol H/ = 1.67mol H ( g N)/(14.01 g/mol) = mol N/ = 1mol N ( g O)/(16 g/mol) = mol O/ = 2 mol O multiply by 3 to get the empirical formula ⇒ C 7 H 5 N 3 O 6 Ch. 3 – Answer Key

38
11. Consider the following unbalanced reaction: 4 NH O 2 4 NO H 2 O a. How many moles of oxygen gas are required to make 12.8 moles of nitrogen dioxide? 12.8 mol NO 2 7 mol O 2 = 22.4 mol O 2 b. How many grams of water can be produced from 9.64 g of ammonia? 9.64 g NH 3 1mol NH 3 6 mol H 2 O g H 2 O = 15.3 g H 2 O Ch. 3 – Answer Key 4 mol NO g NH 3 4 mol NH 3 1 mol H 2 O Continue to next slide…

39
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 NH 3 )/4 = 0.75 vs. (5 mol O 2 )/7 = 0.71 ⇒ O 2 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 NH 3 )/( g/mol) = mol NH 3 /4 = vs. (28 g O 2 )/(32 g/mol) = mol O 2 /7 = ⇒ O 2 is the LR Ch. 3 – Answer Key

40
12. Phosphorus can be prepared from calcium phosphate by the following unbalanced reaction: Ca 3 (PO 4 ) 2 + SiO 2 + C CaSiO 3 + P 4 + CO a. How many grams of carbon monoxide can be produced from a mixture of 10g of each reactant? First ⇒ balance the reaction 2 Ca 3 (PO 4 ) SiO C 6 CaSiO 3 + P CO Next ⇒ determine the limiting reagent (10 g Ca 3 (PO 4 ) 2 )/( g/mol) = mol Ca 3 (PO 4 ) 2 /2 = (10 g SiO 2 )/(60.09 g/mol) = mol SiO 2 /6 = (10 g C)/(12.01g/mol) = mol C/10 = Ca 3 (PO 4 ) 2 is the LR mol Ca 3 (PO 4 ) 2 10 mol CO g CO = 4.51 g CO (theoretical yield) 2 mol Ca 3 (PO 4 ) 2 1 mol CO

41
Ch. 3 – Answer Key 12. …continued b. What is the percent yield if 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 mol Ca 3 (PO 4 ) 2 6 mol SiO g SiO 2 = 5.80 g SiO 2 is consumed mol Ca 3 (PO 4 ) 2 10 mol C g C = 1.9 g C is consumed 2 mol Ca 3 (PO 4 ) 2 1 mol SiO 2 10g – 5.8g = 4.2g SiO 2 are leftover 2 mol Ca 3 (PO 4 ) 2 1 mol C 10g – 1.9g = 8.1g C are leftover

42
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? P 4 + F 2 PF 3 (unbalanced) First ⇒ balance reaction ⇒ P F 2 4 PF 3 Determine theoretical yield from the actual and the % yield ⇒ 63.2% = (83 g/Theo)x100 ⇒ Theoretical yield = g PF 3 Use theoretical yield to determine grams of F g PF 3 1 mol PF 3 6 mol F 2 38 g F 2 = 85.1 g F g PF 3 4 mol PF 3 1 mol F 2

Similar presentations

© 2016 SlidePlayer.com Inc.

All rights reserved.

Ads by Google