1. Unit 1: Stoichiometry
Chemistry 2202
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2. Stoichiometry
Stoichiometry deals with quantities used in OR produced by a chemical reaction
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3. 3 Parts Mole Calculations (Chp. 2 & 3)
Stoichiometry and Chemical Equations (Chp. 4)
Solution Stoichiometry (Chp. 6)
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4. PART 1 - Mole Calculations
Isotopes and Atomic Mass (pp )
Avogadro’s number (pp. 47 – 49)
Mole Conversions (pp )
M, VSTP, NA, n, m, v, N
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5. Questions p. 54 #’s 5 – 8 p. 45 #’s 1 – 4 p. 65 #’s 2, 4, 5
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6. PART 1 - Mole Calculations
Percent composition:
- given mass (p )
- given the chemical formula (p )
Empirical Formulas (pp )
Molecular Formulas (pp )
Lab: Formula of a Hydrate
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7. Questions p. 103 #’s 23 – 24 p. 82 #’s 1 – 4 p. 86 #’s 1, 3 – 6
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8. Isotopes and Atomic Mass
atomic number - the number of protons in an atom or ion
mass number - the sum of the protons and neutrons in an atom
isotope - atoms which have the same number of protons and electrons but different numbers of neutrons
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9. Isotopes and Atomic Mass
eg.
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10. Isotopes and Atomic Mass
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11. Isotopes and Atomic Mass
11 %
79 %
10 %
not all isotopes are created equal
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12. Isotopes and Atomic Mass
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13. Isotopes and Atomic Mass
atomic mass unit (AMU - p.43)
a unit used to describe the mass of individual atoms
the symbol for the AMU is u
1 u is 1/12 of the mass of a carbon-12 atom
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14. Isotopes and Atomic Mass
average atomic mass (AAM)
the AAM is the weighted average of all the isotopes of an element (p. 45)
p. 14 # 5
p. 45 #’s 1 – 3
p. 46 #’s 1 – 4
p. 75 #’s
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15. Finding % Abundance
eg. Br has two naturally occurring isotopes. Br-79 has a mass of u and Br-81 has a mass of u. If the AAM of Br is u, determine the percentage abundance of each isotope.
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16. Finding % Abundance Let y = fraction of Br-81 x + y = 1 y = 1 - x
Let x = fraction of Br-79
Let y = fraction of Br-81
x + y = 1
78.92x y = 79.90
y = 1 - x
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17. x + y = 1
78.92x y = 79.90
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18. Avogadro’s Number (p. 47)
p. 48
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19. Avogadro’s Number
The MOLE is the number of atoms contained in exactly 12 g of carbon-12.
1 mole = x 1023 particles
1 mol = x 1023 particles
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20. NA = 6.022 x 1023 particles/mol Avogadro’s Number
In honor of Amedeo Avogadro, the number of particles in 1 mol has been called Avogadro’s number.
NA = x 1023 particles/mol
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21. N = n x NA Avogadro’s Number n = # of moles Formulas:
N = # of particles (atoms, ions, molecules, or formula units)
NA = Avogadro’s #
Formulas:
N = n x NA
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22. Avogadro’s Number How many moles are contained in the following?
2.56 x Pb atoms
7.19 x CO2 molecules
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23. eg. Calculate the number of moles in 4.98 x 1025 atoms of Al.
Avogadro’s Number
eg. Calculate the number of moles in x 1025 atoms of Al.
eg. How many formula units of Na2SO4 are in 5.69 mol of Na2SO4?
# of Na ions?
# of Oxygen atoms?
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24. Avogadro’s Number
eg. How many molecules of glucose are in mol of C6H12O6?
How many carbon atoms?
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25. Avogadro’s Number
eg. Calculate the number of moles in a sample of glucose that has 3.56 x 1022 glucose molecules.
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26. Avogadro’s Number
pp. 51 – 53: #’s 5 – 15
p. 54:  #’s 4 - 8
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27. Molar Mass
The mass of one mole of a substance is called the molar mass of the substance
eg.
1 mole of Pb has a mass of g
1 mole of Ag has a mass of g
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28. Molar Mass
The symbol for molar mass is M and the unit is g/mol
eg. MPb = g/mol
MAg = g/mol
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29. Molar Mass
The molar mass of a compound is the sum of the molar masses of the elements in the compound
eg. Calculate the molar mass of:
a) H2O b) C6H12O6 c) Ca(OH)2
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30. Molar Mass H2O has 2 hydrogens and 1 oxygen 2 x 1.01 = 2.02
18.02 g/mol
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31. Molar Mass
C6H12O6
6 x = 72.06
12 x 1.01 = 12.12
6 x = 96.00
g/mol
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32. Molar Mass Your calculator may not show Ca(OH)2 the zeroes.
There should be 2 digits after the decimal when adding molar masses
Ca(OH)2
1 x = 40.08
2 x = 32.00
2 x = 2.02
74.10 g/mol
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33. Molar Mass p. 57: #’s 16 – 19 & Molar Masses Handout
g/mol g/mol
g/mol g/mol
g/mol g/mol
g/mol g/mol
g/mol g/mol
g/mol
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34. Molar Mass Calculations
Avogadro’s #
molar mass
N = n x NA
m = n x M
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35. Molar Mass Calculations
m = n x M
N = n x NA
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36. Molar Mass Calculations
eg. How many moles are in g of NO2?
m = 25.3 g
MNO2 = g/mol
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37. Molar Mass Calculations
eg. What is the mass of 4.69 mol of water?
n = 4.69 mol
Mwater = g/mol
m = n x M
= (4.69 mol)(18.02 g/mol)
= g
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38. Molar Mass Calculations
Practice: p #’s
p #’s
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39. Particle–Mole-Mass Conversions
particles
(N)
Moles
(n)
Mass
(m)
5.98 x 1026 Cu atoms
4.50 g H2O
6.15 mol O3
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40. Molar Mass Calculations
x NA
x M
particles(N)
moles (n)
mass (m)
÷ M
÷ NA
Practice: p #’s 28
p #’s 34
p # 15
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41. Molar Mass Calculations
eg. How many molecules are in 26.9 g of water?
m = 26.9 g
Mwater= g/mol
NA = x 1023 molecules/mol
Find N
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42. Molar Mass Calculations
N = n x NA
= X x 1023
= 8.99 x 1023 molecules
= mol H2O
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43. Molar Mass Calculations
eg. How many molecules are in 4.78 g of glucose?
m = 4.78 g
Mwater= g/mol
NA = x 1023 molecules/mol
Find N
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44. Molar Mass Calculations
N = n x NA
= X x 1023
= 8.99 x 1023 molecules
= mol H2O
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45. Molar Mass Calculations
eg. A sample of Sn contains 4.69 x 1028 atoms. Calculate its mass.
N = 4.69 x 1028
NA = x 1023 molecules/mol
MSn = g/mol
Find m
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46. Molar Mass Calculations
m = n x M
= mol x g/mol
= 9.24 x 10 6 g
= 77,881 mol
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47. Molar Mass Calculations
Practice: p #’s
p #’s 34 – 37
p # 15
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48. Molar Mass Calculations
Practice:
p #’s 5 - 8
p #’s 2, 4, 5
p #’s 13, 14,
p #’s 15, 17 – 19,
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49. Molar Volume Gases are frequently measured using volume in litres (L).
Gas volume increases when temperature increases but decreases when pressure increases .
The volume of gases is measured under conditions of Standard Temperature and Pressure (STP)
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50. Molar Volume Standard Pressure – 101.3 kPa Standard Temperature – 0 °C
Avogadro hypothesized that equal volumes of gases at the same temperature and pressure contain equal numbers of molecules.
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51. Molar Volume
Experimental evidence shows the volume of one mole of ANY GAS at STP is 22.4 L/mol
OR VSTP = 22.4 L/mol
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52. v = n x VSTP Molar Volume given volume in Litres N = n x NA m = n x M

53. Sample Questions Calculate the number of moles in 36.4 L
of CO2 gas at STP.
v = 36.4 L
VSTP = 22.4 L/mol
Find n
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54. Sample Questions Calculate the volume of 0.479 mol of
helium gas at STP.
n = mol
VSTP = 22.4 L/mol
Find v
v = n x VSTP
= mol x 22.4 L/mol
= 10.7 L
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55. Mole Calculations x NA x M particles(N) moles (n) mass (m) ÷ M ÷ NA
÷ VSTP
x VSTP
volume (v)
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56. Sample Questions 2 steps Find v v = n x VSTP
Calculate the volume of 86.4 g of CO2 gas at STP.
m = 86.4 g
M = g/mol
VSTP = 22.4 L/mol
Find v
v = n x VSTP
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57. Sample Questions 2 steps Find m
Calculate the mass of L of propane gas, C3H8, at STP.
v = L
VSTP = 22.4 L/mol
M = g/mol
Find m
m = n x M
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58. Molar Volume p. 73 #’s 38 – 43 p. 74 #’s 1 – 4 p. 76 #’s 26, 27
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59. 10.0 g of an oxide of nitrogen has a volume
Extra:
10.0 g of an oxide of nitrogen has a volume
of 7.46 L at STP. Determine whether the formula
of the oxide is NO, N2O or NO2.
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60. Percent Composition (p. 79)
The mass percent of a compound is the mass of an element expressed as a percent of the total mass of the compound.
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61. Percent Composition
eg g of a compound was analyzed and found to contain 6.00 g of hydrogen and 2.50 g of carbon. Calculate the mass percent for each element.
p. 82 #’s 1 - 4
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62. Percent Composition
mass percent may be found using the formula & the molar mass of a compound.
eg.
Find the percentage composition for CH4
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63. Percent Composition M = 12.01 g/mol + 4(1.01 g/mol)
% H = (4.04/16.05) X 100% = 25.2 %
% C = (12.01/16.05) X 100 % = 74.8 %
p. 85 #’s 5 - 8
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64. p. 86 #’s 1, 3 – 6
p #’s 6 – 10
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65. Empirical Formulas
An empirical formula gives the simplest ratio of elements in a compound.
A molecular formula shows the actual number of atoms in a molecule of a compound.
Ionic compounds are always written as empirical formulas
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66. Empirical Formulas Compound Molecular Formula Empirical Formula butane
C4H10
glucose
C6H12O6
water
H2O
benzene
C6H6
C2H5
CH2O
H2O CH
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67. Empirical Formulas (p.87)
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68. Empirical Formulas
The empirical formula of a compound may be determined by using the % composition of a given compound.
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69. Empirical Formulas Method:
assume you have g of the compound (ie. change % to g)
calculate the moles (n) for each element
divide each n by the smallest n to get the ratio for the empirical formula
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70. Empirical Formulas
eg. A compound was analyzed and found to contain 87.4% N and 12.6 % H by mass. Determine the empirical formula of the compound.
p. 89 #’s
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71. Empirical Formulas
When finding the EF, the mole ratio may not be a whole number ratio.
eg. A compound contains 84.73% N and % H by mass. Determine the empirical formula of the compound.
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72. p. 90
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73. Empirical Formulas
eg. A compound contains 89.91% C and % H by mass. Determine the empirical formula of the compound.
p. 91 #’s 13 – 16
p. 94 #’s 2-4, 6, 7
Answers on p. 109
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74. MgO Lab
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75. Molecular Formulas
The molecular formula of a compound is a multiple of the empirical formula.
See p. 95
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76. Molecular Formulas
To find the molecular formula we need the empirical formula and the molar mass of the compound
eg. The empirical formula of hydrazine is NH2. The molar mass of hydrazine is g/mol. What is the molecular formula for hydrazine?
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77. Molecular Formulas
eg. A hydrocarbon contains 85.6% carbon and 14.4% hydrogen by mass. The molar mass of the compound is g/mol. Determine the molecular formula for the hydrocarbon?
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78. Molecular Formulas
p #’s 17 – 20
p. 107, 108 #’s
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79. CHC analyzer (p. 99 – 101)
1. Describe the operation of a carbon hydrogen combustion analyzer.
g of carbon dioxide and 10.8 g of water is collected in a CHC analysis. Determine the empirical formula of the hydrocarbon.
p #’s 21, 22
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80. Formula of a hydrate To determine the formula of a hydrate:
- calculate the moles of water
- calculate the moles of anhydrous compound
- determine the simplest ratio
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81. Formula of a hydrate
eg. Use the data below to determine the value of x in LiCl• xH2O.
mass of crucible = g
crucible + hydrate = g
crucible + anhydrous compound= g
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82. mwater = – = g H2O
mLiCl = – = g LiCl
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83. eg. Na2CO3. xH2O crucible = 15.96 g crucible + hydrate = 22.19 g
crucible + anhydrous compound = g
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84. crucible + anhydrous compound = 158.23 g p. 103; # 24 Lab: pp. 104-105
eg. CoCl2.xH2O
crucible = g
crucible + hydrate = g
crucible + anhydrous compound = g
p. 103; # 24 Lab: pp
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85. Formula of a hydrate mass of empty beaker mass of beaker & hydrate
mass of beaker & anhydrous compound
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86. Review – Chp. 3 p. 86 #’s 1, 3 – 6 p. 94 #’s 1 – 7 p. 103 # 23
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87. Test p. 54 #’s 5 – 8 p. 45 #’s 1 – 4 p. 65 #’s 2, 4, 5 p. 46 #’s 1 – 6
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88. Test
p. 82 #’s 1 – 4
p. 85 #’s 5 – 8
p. 89 #’s 9 – 12
p. 91 #’s 13 – 16
p. 97 #’s
p. 103 #’s 23 – 24
p. 86 #’s 1, 3 – 6
p. 94 #’s 1 - 7
p. 106 #’s 1 - 3, 6, 7
p. 107 – 109
#’s 5 – 23, 25
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89. Stoichiometry (Chp.4)
Stoichiometry is the determination of quantities needed for, or produced by, chemical reactions.
Ratios from balanced chemical equations are used to predict quantities.
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90. Clubhouse sandwich recipe
Stoichiometry – p. 111
Clubhouse sandwich recipe
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91. Clubhouse sandwich recipe
Fill in the missing quantities:
Slices of Toast
Slices of Turkey
Strips of Bacon
# of Sandwiches 12 27 66
100
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92. Mole Ratios
A mole ratio is a mathematical expression that shows the relative amounts of two species involved in a chemical change.
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93. Mole Ratios A mole ratio Come from a balanced chemical equation
Shows the relative amounts of the reactants/products in moles
Is the coefficient for the required species in the numerator and the coefficient for the given species in the denominator.
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94. N2(g) + 3 H2 (g) → 2 NH3 (g) 20 66
140 81
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95. C3H8 + 5 O2 → 3 CO2 + 4 H2O How many moles of CO2 are
produced when 31.5 mol of O2 react?
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96. C3H8 + 5 O2 → 3 CO2 + 4 H2O 2. How many moles of H2O are produced
when 1.35 mol of O2 react?
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97. C3H8 + 5 O2 → 3 CO2 + 4 H2O 3. How many moles of O2 are needed to
produce 31.5 mol of CO2?
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98. C3H8 + 5 O2 → 3 CO2 + 4 H2O 4. How many moles of C3H8 are needed to
react with mol of O2?
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99. C5H12 + O2 → CO2 + H2O How many moles of CO2 are produced
when 6.35 mol of O2 react?
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100. Al(s) + Br2(l) → AlBr3(s) How many moles of Br2 are needed to
produce mol of AlBr3?
p #’s 4 – 7
p #’s
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101. Mole to Mole Stoichiometry
How many moles of nitrogen gas are needed to produce 6.75 mol of NH3 in a reaction with hydrogen gas?
How many moles of silver would be produced if 10.0 mol of silver nitrate reacts with copper metal?
How many moles of water are produced when 20.6 mol of CH4 burns?
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102. Mole to Mole Stoichiometry
How many moles of copper would be produced if 20.5 mol of copper (II) oxide decomposes?
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103. Mole to Mole Stoichiometry
How many moles of copper would be produced if 20.5 mol of copper (II) oxide decomposes?
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104. Mass to Mole Stoichiometry
How many moles of water are produced when 20.6 g of CH4 burns?
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105. Mass to Mole Stoichiometry
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106. Mole to Mass Stoichiometry
What mass of CaCl2 is produced when
4.38 mol of Ca(NO3)2 reacts with NaCl?
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107. 9:53 PM
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108. Mole to Mass Stoichiometry
What mass of CaCl2 is produced when
4.38 mol of Ca(NO3)2 reacts with NaCl?
How many moles of copper would be produced if 20.5 g of copper (II) oxide decomposes?
How many moles of water are produced when 5.45 gl of C3H8 burns?
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109. Mass to Mass Stoichiometry
eg. Calculate the mass of HCl needed to react with 3.56 g of Fe to produce FeCl2.
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110. Ca(NO3)2 + 2 NaCl → CaCl2 + 2 NaNO3
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111. 2 CuO → 2 Cu + O2
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112. C3H O2 → 3 CO H2O
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113. Stoichiometry (Chp.4) Four step stoichiometry:
Write a balanced chemical equation
Calculate moles given from mass
Mole ratio – find moles required
Calculate required quantity (mass)
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114. Stoichiometry (Chp.4)
eg. What mass of CO2 gas is produced when 45.9 g of CH4 burns ?
Step #1
CH O2 → CO H2O
45.9 g ? g
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115. 9:53 PM

116. 9:53 PM

117. eg. How many moles of HCl needed to react with 3
eg. How many moles of HCl needed to react with 3.56 g of Fe to produce FeCl2.
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118. Mole Calculations (p. 121 #13)
3.56 g
? g
Fe HCl → FeCl H2
Step #2
= mol Fe
Step #3
= mol HCl
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119. Mole Calculations p. 122 #15 Given 32.0 g of sulfur (M = 256.56 g/mol)
Find mass of ZnS
#2 n = mol S8
#3 n = mol ZnS
(M = g/mol)
#4 m = 97.2 g ZnS
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120. Mole Calculations p. 123 #18 Given 33.5 g of H3PO4 (M = 98.00 g/mol)
Find mass of MgO
#2 n = mol H3PO4
#3 n = mol MgO
(M = g/mol)
#4 m = 20.7 g MgO
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121. Mole Calculations p. 123 #17 Given 25.0 g of Al4C3 (M = 143.95 g/mol)
Find volume of CH4
#2 n = mol Al4C3
#3 n = mol CH4
(MV = 22.4 L/mol)
#4 m = 11.7 L CH4
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122. How many moles of aluminum chloride can be produced from the reaction of chlorine and 10.8 mol of aluminum ?
Cl2(g) + Al(s) → AlCl3(s)
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123. How many moles of magnesium are needed to react with 27 g of iodine to form magnesium iodide?
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124. How many grams of nitrogen are needed to react with 14
How many grams of nitrogen are needed to react with 14.0 mol of oxygen to produce nitrogen dioxide ?
N2(g) O2(g) → NO2(g)
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125. Mole Calculations (p. 121 #14)
2.34 g
? L
#2 n = mol NO2
#3 n = mol O2
(MV = 22.4 L/mol)
#4 n = ( )(22.4) = L O2
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126. Limiting Reactant (p. 128) 10.0 g of Li requires _______ of Br2
15.0 g of Br2 requires _______ of Li
10.0 g of Li produces _______ of LiBr
15.0 g of Br2 produces ______ of LiBr
This problem has _____ g of excess _____
and will produce ______ g of LiBr
Half the class finds yield from 10.0 g of Li.
Other half find yield from 15.0 g of bromine.
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126

127. Limiting Reactant (p. 128)
10.0 g of Li reacts with 15.0 g of Br2. Calculate the mass of LiBr produced.
Half the class finds yield from 10.0 g of Li.
Other half find yield from 15.0 g of bromine.
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128. Limiting Reactant (p. 128)
The Limiting Reactant (LR) OR Limiting Reagent (LR) is the substance that is completely used in a chemical reaction.
The Excess Reactant is the reactant that is left over after a reaction is complete.
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129. Limiting Reactant (p. 128)
eg g of NaI reacts with 2.00 g of Pb(NO3)2. Determine the LR and calculate the amount of PbI2 produced.
write a balanced equation
find n for each reactant (Step #2)
find moles produced by each reactant (Step #3)
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130. Pb(NO3)2 + 2 NaI → 2 NaNO3 + PbI2 2.00 g 2.00 g nPb(NO)3 = 2.00 g
g/mol
= mol
nNaI = g
g/mol
= mol
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131. nPbI2 = 0.006038 mol Pb(NO3)2 x 1 mol PbI2 1 mol Pb(NO3)2
nPbI2 = mol NaI x 1 mol PbI2
2 mol NaI
= mol PbI2
mPbI2 = mol x g/mol
= g PbI2
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132. nCaCO3 = 0.06448 mol Ca3(PO4)2 x 3 mol CaCO3 1 mol Ca3(PO4)2
What mass of calcium carbonate will be produced when 20.0 g of calcium phosphate reacts with 15.0 g of sodium carbonate? (14.2 g)
nCa3(PO4)2 = g
g/mol
= mol
nNa2CO3 = g
g/mol
= mol
nCaCO3 = mol Ca3(PO4)2 x 3 mol CaCO3
1 mol Ca3(PO4)2
= mol CaCO3
nCaCO3 = mol Na2CO3 x 3 mol CaCO3
3 mol Na2CO3
= mol CaCO3
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133. What mass of barium hydroxide will be produced when 10
What mass of barium hydroxide will be produced when 10.0 g of barium nitrate reacts with 30.0 g of sodium hydroxide? (6.56 g)
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134. What volume of hydrogen gas at STP will be produced when 10
What volume of hydrogen gas at STP will be produced when 10.0 g of zinc metal reacts with 20.0 g of hydrogen chloride?
Zn HCl → H2 + ZnCl2
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135. Ba(NO3)2 + 2 NaOH → 2 NaNO3 +Ba(OH)2
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136. 9:53 PM

137. Law of Conservation of Mass (p. 118)
In a chemical reaction, the total mass of reactants always equals the total mass of products.
eg. 2 Na3N → 6 Na + N2
When g of Na3N decomposes g of N2 is produced. How much Na is produced in this decomposition?
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138. Law of Conservation of Mass ( p. 118)
eg. To produce 90.1 g of water, what mass
of hydrogen gas is needed to react with
80.0 g of oxygen?
eg. If 3.55 g of chlorine reacts with exactly
2.29 g of sodium, what mass of NaCl will
be produced?
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139. Percent yield (p. 137)
The theoretical yield is the amount of product that we calculate using stoichiometry
The actual yield is the amount of product obtained from a chemical reaction
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140. Percent yield (p. 137)
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141. DEMO: silver nitrate + copper
Equation:
Mass AgNO3 =
Mass Cu =
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142. DEMO: silver nitrate + copper
Mass of filter paper and precipitate =
Mass of empty filter paper =
Mass of precipitate =
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