tis the season to be thankful so lets thank avogadro for math in chemistry
TRANSCRIPT
![Page 1: Tis the season to be thankful so lets thank Avogadro for math in Chemistry](https://reader036.vdocuments.us/reader036/viewer/2022062720/56649eff5503460f94c1545e/html5/thumbnails/1.jpg)
Tis the season to be thankful so lets thank Avogadro for math in
Chemistry
![Page 2: Tis the season to be thankful so lets thank Avogadro for math in Chemistry](https://reader036.vdocuments.us/reader036/viewer/2022062720/56649eff5503460f94c1545e/html5/thumbnails/2.jpg)
Percent Composition
Purpose: Can be used to figure out chemical formulas. the percentage by mass of each element in a
compound
100mass total
element of massncompositio %
![Page 3: Tis the season to be thankful so lets thank Avogadro for math in Chemistry](https://reader036.vdocuments.us/reader036/viewer/2022062720/56649eff5503460f94c1545e/html5/thumbnails/3.jpg)
I. Percent Composition
Two different types of problems: 1) Masses are given 2) No Masses are given
![Page 4: Tis the season to be thankful so lets thank Avogadro for math in Chemistry](https://reader036.vdocuments.us/reader036/viewer/2022062720/56649eff5503460f94c1545e/html5/thumbnails/4.jpg)
Masses are Given
Steps to solve problem: 1) Add given masses to get total mass for one
compound 2) Divide mass of each element by the total mass 3) Multiply by 100 to get the percent
![Page 5: Tis the season to be thankful so lets thank Avogadro for math in Chemistry](https://reader036.vdocuments.us/reader036/viewer/2022062720/56649eff5503460f94c1545e/html5/thumbnails/5.jpg)
%Fe =28 g
36 g 100 =78% Fe
%O =8.0 g
36 g 100 =22% O
Find the percentage composition of a sample that is 28 g Fe and 8.0 g O.
A. Percentage Composition
![Page 6: Tis the season to be thankful so lets thank Avogadro for math in Chemistry](https://reader036.vdocuments.us/reader036/viewer/2022062720/56649eff5503460f94c1545e/html5/thumbnails/6.jpg)
No Masses Given
Steps to solve problem: 1) Assume you have 1 mole of the compound 2) Calculate the molar mass of each element in
the compound by multiplying the subscript by the molar mass of the element
3) Divide the molar mass for the element by the total molar mass of the compound
4) Multiply by 100 to get the percent
![Page 7: Tis the season to be thankful so lets thank Avogadro for math in Chemistry](https://reader036.vdocuments.us/reader036/viewer/2022062720/56649eff5503460f94c1545e/html5/thumbnails/7.jpg)
No Masses GivenExamples
1) Calculate the percent composition of oxygen in water.
H2O H: 2 x 1.008 = 2.016 g/mol O: 1 x 16.00 = 16.00 g/mol Total molar mass (2.016 + 16.00) = 18.02
g/mol %O = (16.00 / 18.02) x 100 = 88.79%
![Page 8: Tis the season to be thankful so lets thank Avogadro for math in Chemistry](https://reader036.vdocuments.us/reader036/viewer/2022062720/56649eff5503460f94c1545e/html5/thumbnails/8.jpg)
No Masses GivenExamples
2) Calculate the percent composition of calcium carbonate.
100.09 g/mol %Ca = 40.08g/100.09g×100 = 40.04% %C = 12.01g/100.09g×100 =12.00% %O = 3(16.00g)/100.09g×100 = 47.96%
![Page 9: Tis the season to be thankful so lets thank Avogadro for math in Chemistry](https://reader036.vdocuments.us/reader036/viewer/2022062720/56649eff5503460f94c1545e/html5/thumbnails/9.jpg)
B. Empirical Formula
C2H6
CH3
reduce subscripts
Smallest whole number ratio of atoms in a compound
![Page 10: Tis the season to be thankful so lets thank Avogadro for math in Chemistry](https://reader036.vdocuments.us/reader036/viewer/2022062720/56649eff5503460f94c1545e/html5/thumbnails/10.jpg)
B. Empirical Formula1. Find mass (or %) of each element.
2. Find moles of each element. (divide given mass by molar mass)
3. Divide answers by the smallest # to find subscripts.
4. When necessary, multiply subscripts by 2, 3, or 4 to get whole #’s.
![Page 11: Tis the season to be thankful so lets thank Avogadro for math in Chemistry](https://reader036.vdocuments.us/reader036/viewer/2022062720/56649eff5503460f94c1545e/html5/thumbnails/11.jpg)
Remember!
Percent to MassMass to MoleDivide by SmallMultiply til Whole
![Page 12: Tis the season to be thankful so lets thank Avogadro for math in Chemistry](https://reader036.vdocuments.us/reader036/viewer/2022062720/56649eff5503460f94c1545e/html5/thumbnails/12.jpg)
B. Empirical FormulaFind the empirical formula for a
sample of 25.9% N and 74.1% O.
25.9 g 1 mol
14.01 g = 1.85 mol N
74.1 g 1 mol
16.00g = 4.63 mol O
1.85 mol
1.85 mol
= 1 N
= 2.5 O
![Page 13: Tis the season to be thankful so lets thank Avogadro for math in Chemistry](https://reader036.vdocuments.us/reader036/viewer/2022062720/56649eff5503460f94c1545e/html5/thumbnails/13.jpg)
B. Empirical Formula
N1O2.5
Need to make the subscripts whole numbers multiply by 2
N2O5If .0_ something or .9_ something, just round
![Page 14: Tis the season to be thankful so lets thank Avogadro for math in Chemistry](https://reader036.vdocuments.us/reader036/viewer/2022062720/56649eff5503460f94c1545e/html5/thumbnails/14.jpg)
Empirical FormulasExamples
3) Ascorbic acid (vitamin C) contains C (40.89%), H (4.56%), and O (54.55%) by mass. What is the empirical formula of ascorbic acid?
C3H4O3
![Page 15: Tis the season to be thankful so lets thank Avogadro for math in Chemistry](https://reader036.vdocuments.us/reader036/viewer/2022062720/56649eff5503460f94c1545e/html5/thumbnails/15.jpg)
C. Molecular Formula“True Formula” - the actual number
of atoms in a compound
CH3
C2H6
empiricalformula
molecularformula
?
![Page 16: Tis the season to be thankful so lets thank Avogadro for math in Chemistry](https://reader036.vdocuments.us/reader036/viewer/2022062720/56649eff5503460f94c1545e/html5/thumbnails/16.jpg)
IV. Molecular Formulas
Usually the empirical formula is the molecular formula for a compound. When it is not, the molecular formula is defined as the elements and number of atoms that are contained in a compound.
![Page 17: Tis the season to be thankful so lets thank Avogadro for math in Chemistry](https://reader036.vdocuments.us/reader036/viewer/2022062720/56649eff5503460f94c1545e/html5/thumbnails/17.jpg)
IV. Molecular Formulas
Molecular formulas are always multiples of empirical formulas.
CH3 C2H6
![Page 18: Tis the season to be thankful so lets thank Avogadro for math in Chemistry](https://reader036.vdocuments.us/reader036/viewer/2022062720/56649eff5503460f94c1545e/html5/thumbnails/18.jpg)
C. Molecular Formula1. Find the empirical formula.2. Find the empirical formula mass.3. Divide the molecular mass by the empirical mass.4. Multiply each subscript by the answer from step 3.
nmass EF
mass MF nEF
![Page 19: Tis the season to be thankful so lets thank Avogadro for math in Chemistry](https://reader036.vdocuments.us/reader036/viewer/2022062720/56649eff5503460f94c1545e/html5/thumbnails/19.jpg)
C. Molecular FormulaThe empirical formula for ethylene is CH2. Find the molecular
formula if the molecular mass is 28.0532 g/mol?
28.0532 g/mol
14.03 g/mol = 2.000
empirical mass = 14.03 g/mol
(CH2)2 C2H4
![Page 20: Tis the season to be thankful so lets thank Avogadro for math in Chemistry](https://reader036.vdocuments.us/reader036/viewer/2022062720/56649eff5503460f94c1545e/html5/thumbnails/20.jpg)
Molecular FormulasExamples
2) The compound methyl butanoate smells like apples. Its percent composition is 58.8% C, 9.8% H, and 31.4% O. If the molecular mass is 102.1317 g/mol, what is the molecular formula?
![Page 21: Tis the season to be thankful so lets thank Avogadro for math in Chemistry](https://reader036.vdocuments.us/reader036/viewer/2022062720/56649eff5503460f94c1545e/html5/thumbnails/21.jpg)
Molecular FormulasExamples
3) You find 7.36 g of a compound has decomposed to give 6.93 g of oxygen. The rest is hydrogen. If the molecular mass is 34.0147 g/mol, what is the molecular formula?