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Get out your notes and locate the following problem:
896 dL of CO2 gas contains how many atoms?
2.4e24
Day 4 10-3
Activity Series LabDay 4 10-3
Propose an activity series based on the following results:
A + BY AY + B
A + CY NO RXN
Day 5 10-4
A > B
C > A
C AB
CHAPTER QUIZ FRIDAY 10-7!!!
Day 5 10-4
Read pages 360-361 AND complete # 15 on page 361.
Day 4 10-3
Fe(s) + Pb(NO3)2(aq) Fe(NO3)2(aq) + Pb(s)
Ca(s) + 2H2O(l) H2(g) + Ca(OH)2(aq)
Cl2(aq) + 2NaI(aq) 2NaCl(aq) + I2(aq)
Zn(s) + H2SO4(aq) ZnSO4(aq) + H2(g)
Percentage Composition
Mass of element Mass of compound
X (100) =
% element in compound
… tells how much an element contributes to the mass of the compound
Percentage Composition
H2O???
2 H: 2 * 1.0079 g = 2.0158 g
1 O: 1 * 15.999 g = 15.999 g
18.015 g
(2.0158 g / 18.015) X 100 = 11.190% H
(15.999 g / 18.015) X 100 = 88.810% O
Percentage CompositionAn unknown compound w/ a mass of 0.237 g is extracted from the roots of a plant. Decomposition of the sample produces 0.0948 g of C, 0.1264 g of O, and 0.0158 g of H. What is the % composition of the compound?
Day 6 10-5
0.112 kL of CO gas contains how many grams? How many atoms?
140 grams CO
6e24 atoms
Review (if needed) pages 325-327 AND complete #s 33, 34, 35, and 36 on pages 326 and 327.
Day 5 10-4
Empirical formula – smallest whole-number mole ratio for a compound
Empirical Formulas
To Calculate:1. Convert all elements involved to
moles
2. Divide all elements by the smallest # of moles
3. Obtain smallest whole #ed ratio
A compound contains 13.5 g Ca, 10.8 g O, and 0.675 g H. What is the empirical formula?
Empirical Formulas
1. Convert all eles. involved to moles
2. Divide all eles. by the smallest # of moles
3. Obtain smallest whole #ed ratio
Ca = 0.337 moles
O = 0.675 moles
H = .675 moles
Ca = 0.337 mols / 0.337 mols = 1
O = 0.675 mols / 0.337 mols = 2
H = 0.675 mols / 0.337 mols = 2
1 Ca : 2 O : 2 H CaO2H2Ca(OH)2
Calculating Empirical Formulas
In an unknown compound you find 4.04 g of N and 11.46 g O. Empirical formula?
67.2 L of CH4 gas = ___ grams
44.8 mg of solid Carbon = ___ L
Day 1 10-6
Homework # 2 = now
Postlabs = Friday 10-7 (tomorrow)
Presentations = Wednesday 10-12
Chapter Quiz = Wednesday 10-12
Day 1 10-6
Empirical formula – smallest whole-number mole ratio for a compound
Empirical Formulas vs. Molecular Formulas
Molecular formula – actual # of atoms of each ele. in a molecular compound
Sometimes But not always! the same.
Molecular Formulas
To Calculate:
Compare the molar mass of the empirical formula to the molar mass of the molecular formula
Molecular formula – actual # of atoms of each ele. in a molecular compound
Determine the molecular formula of the compound with an empirical formula of CH and a formula mass of 78.110 amu
Empirical mass = 13.019 g/mol
x = 6
Molecular formula = C6H6
Determining Molecular Formulas
Determining Molecular Formulas
In an unknown compound you find 4.04 g of N and 11.46 g O. Empirical formula? This unknown compound has a molar mass of 108.0 g/mol. What is the molecular formula?
Empirical formula = N2O5
Empirical mass = 108.009 g/mol
Molecular formula = N2O5
A sample of a compound contains 6.0 g of C and 16 g of O. The total sample is 22 g. The molecular molar mass is 44 grams
Percentage Composition
Empirical formula
Molecular formula
Determining Molecular Formulas
Review section 10.3 (if needed) and complete #s 45-49
for # 49 the %s can be treated as grams and used to find moles…
Assignment due Tuesday 10-11
Calculate the mass of Cu produced?
Mass of beaker and Cu – mass of beaker
Calculating percent yield
Percent yield – a way to compare how much you “should” get to how much you actually got
Percent Yield = Actual
yieldTheoretic yieldX 100
A sample of a compound contains 6.0 g of C and 16 g of O. The total sample is 22 g.
Percentage Composition
Empirical formula
Determining Molecular Formulas
Pd. 6
- 2 molar mass convs.
- 2 Av.’s # convs.
- 2 Av.’s law convs.
- 2 multi-step convs.
- 1 percentage comp.
- 1 empirical formula
- 1 molecular formula
- 3 content ?s (no math)
14 questions
Mass of element Mass of compound
X (100) =% element in compound
H2O???2 H: 2 * 1.0079 g = ________ g
1 O: 1 * ________ g = _________ g
(__________________) X 100 = _______% H
(___________________) X 100 = _______% O
A compound contains 13.5 g Ca, 10.8 g O, and 0.675 g H. What is the empirical formula?
1. Convert all eles. involved to moles
2. Divide all eles. by the smallest # of moles
3. Obtain smallest whole #ed ratio
Ca = 0.337 moles
O = 0.675 moles
H = .675 moles
Ca = 0.337 mols / 0.337 mols = 1
O = 0.675 mols / 0.337 mols = 2
H = 0.675 mols / 0.337 mols = 2
1 Ca : 2 O : 2 H Ca(OH)2
13.5 g Ca
40 g Ca
1 mol Ca
10.8 g O
16 g O
1 mol O
0.675 g H
1 g H
1 mol H
Percentage Composition
Empirical formula
Molecular formula