1. 294 g of sulfuric acid = ____ molecules 2. How many atoms of O? 1.8 * 10 24 molecules 7.2 * 10 24...
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Transcript of 1. 294 g of sulfuric acid = ____ molecules 2. How many atoms of O? 1.8 * 10 24 molecules 7.2 * 10 24...
![Page 1: 1. 294 g of sulfuric acid = ____ molecules 2. How many atoms of O? 1.8 * 10 24 molecules 7.2 * 10 24 atoms Day 5 10-14.](https://reader036.fdocuments.in/reader036/viewer/2022062314/56649e745503460f94b7533e/html5/thumbnails/1.jpg)
1. 294 g of sulfuric acid = ____ molecules
2. How many atoms of O?
1.8 * 1024 molecules
7.2 * 1024 atoms
Day 5 10-14
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A chemical reaction produces 98.0 mL of sulfur dioxide gas at STP. What was the mass
(in grams) of the gas produced?
0.280 g SO2
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896 dL of CO2 gas contains how many atoms?
2.4e24 molecules
7.2e24 atoms
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Converting:Mass to moles
Moles to mass
Moles to atoms / molecules / particles
Atoms / molecules / particles to moles
Moles to liters
Liters to moles
Changing a substance
=
=
____________________________
______________________
=__________________________
= ________________
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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
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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 g) X 100 = 11.190% H
(15.999 g / 18.015 g) X 100 = 88.810% O
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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?
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1) 45 L of hydrogen gas = how many grams?
Day 6 10-15
2) A sample of a compound contains 12.15 g of Mg and 19 g of F. What is the percent composition of this compound (SHOW WORK)?
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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
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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
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Calculating Empirical Formulas
In an unknown compound you find 4.04 g of N and 11.46 g O. Empirical formula?
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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.
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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
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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
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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
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What result do you expect from a match test if ….
hydrogen is present?
carbon dioxide is present?
Day 1 10-16
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67.2 L of CH4 gas = ___ grams
44.8 mg of solid Carbon = ___ L
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Prelab # 5:
According to the reaction in # 1, for every ___ pieces of cupric chloride ___ pieces of copper are produced. As a result we should expect to produce ___ of a mole of cpper in the first reaction, which would be ___ grams of Cu. SHOW YOUR WORK!
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A sample of a compound contains 3.003 g of C, 0.504 g of H, and 4.000 g of O. If the molar mass is 180.157 g/mol what is the molecular formula.
Empirical formula
Molecular Formula
Question of the Day
Day 2 10-17
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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
Day 3 10-18
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56,000 mL of O2 gas = how many molecules of O2? How many atoms of O?
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- 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
- 2 content ?s (no math)
13 questions
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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
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Book #s 45-49 = NOW
Create-a-Test = momentarily
Presentations = Wednesday 10-12
Chapter Quiz = Wednesday 10-12
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Calculate the mass of Cu produced?
Mass of beaker and Cu – mass of beaker
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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
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Desk clear and mentally prepare for Quest.
Quest
Group Quest
Presentations
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Day 6 10-5
0.112 kL of CO gas contains how many grams? How many atoms?
140 grams CO
6e24 atoms
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Review (if needed) pages 325-327 AND complete #s 33, 34, 35, and 36 on pages 326 and 327.
Day 5 10-4
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Homework # 2 = now
Postlabs = Friday 10-7 (tomorrow)
Presentations = Wednesday 10-12
Chapter Quiz = Wednesday 10-12
Day 1 10-6
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1. A mole is a _________, plus it relates to ______.
2. Molar mass of Be(NO3)2
3. 3.0 grams of Be(NO3)2 = ___ moles
4. = ___ molecules?