The Mole: A Measurement of Matter Describe how Avogadro’s number is related to a mole of any...
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![Page 1: The Mole: A Measurement of Matter Describe how Avogadro’s number is related to a mole of any substance Calculate the mass of a mole of any substance.](https://reader036.fdocuments.in/reader036/viewer/2022062314/56649de85503460f94ae1c5e/html5/thumbnails/1.jpg)
The Mole: A Measurement of Matter
•Describe how Avogadro’s number is related to a mole of
any substance•Calculate the mass of a mole of
any substance
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The Mole and Avogadro’s Number
SI unit that measures the amount of substance
1 mole = 6.022 x 1023 representative particles
Representative particles are usually atoms, molecules, or formula units (ions)
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ExamplesSubstance Representa
tive ParticleChemical Formula
Representative Particles in 1.00 mol
Atomic nitrogen
Atom N 6.022 x 1023
Water Molecule H2O 6.022 x 1023
Calcium ion
Ion Ca2+ 6.022 x 1023
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SolveSubstance Representative
ParticleChemical Formula
Representative Particles in 1.00 mol
Nitrogen gas
Calcium Fluoride
Sucrose
Carbon
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Answers
Nitrogen gas-molecule-N2
Calcium fluoride-formula unit-CaF2
Sucrose-molecule-C12H22O11
Carbon-atom-C
All have 6.022 x 1023 representative particles in 1.00 mol
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How many atoms are in a mole?
Determined from the chemical formula
List the elements and count the atoms
Solve for CO2
C - 1 carbon atom
O - 2 oxygen atoms
Add: 1 + 2 = 3
Answer: 3 times Avogadro’s number of atoms
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Solve: How many atoms are in a mole of
1. Carbon monoxide – CO
2. Glucose – C6H12O6
3. Propane – C3H8
4. Water – H2O
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How many moles of magnesium is 1.25 x 1023 atoms of magnesium?
Divide the number of atoms or molecules given in the example by 6.022 x 1023
Divide (1.25 x 1023) by (6.022 x 1023)
Express in scientific notation
Answer = 2.08 x 10-1 mol Mg
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Objectives
Use the molar mass to convert between mass and moles of a substance
Use the mole to convert among measurements of mass, volume, and number of particles
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Molar mass
Mass (in grams) of one mole of a substanceBroad term (can be substituted) for gram atomic mass, gram formula mass, and gram molecular massCan be unclear: What is the molar mass of oxygen?
O or O2 ? - element O or molecular compound O2 ?
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Molar Mass
Gram atomic mass (gam) – atomic mass of an element taken from the periodic table
Gram molecular mass (gmm) – mass of one mole of a molecular compound
Gram formula mass (gfm) – mass of one mole of an ionic compound
Can use molar mass instead of gam, gmm, or gfm
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Calculating the Molar Mass of Compounds (Molecular and Ionic)
1. List the elements
2. Count the atoms
3. Multiply the number of atoms of the element by the atomic mass of the element (atomic mass is on the periodic table)
4. Add the masses of each element
5. Express to hundredths place
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What is the molar mass (gfm) of ammonium carbonate (NH4)2CO3?
N 2 x 14.01 g = 28.02 g
H 8 x 1.01 g = 8.08 g
C 1 x 12.01 g = 12.01 g
O 3 x 16.00 g = 48.00 g
Add ________
Answer 96.11 g
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Practice Problems1. How many grams are in 9.45 mol of dinitrogen trioxide (N2O3) ?
a. Calculate the grams in one mole b. Multiply the grams by the number of moles
2. Find the number of moles in 92.2 g of iron(III) oxide (Fe2O3).
a. Calculate the grams in one mole b. Divide the given grams by the
grams in one mole
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Answers
1. 718.39 g N2O3 (one mole is 76.02 g)
2. 0.577 mol Fe2O3 (one mole is 159.70 g)
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Volume of a Mole of Gas
Varies with a change in temperature or a change in pressure
At STP, 1 mole of any gas occupies a volume of 22.4 L
Standard temperature is 0°C
Standard pressure is 101.3 kPa (kilopascals), or 1 atmosphere (atm)
22.4 L is known as the molar volume
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22.4 L of any gas at STP contains 6.022 x 1023 representative particles of that gas
One mole of a gaseous element and one mole of a gaseous compound both occupy a volume of 22.4 L at STP (Masses may differ)
Molar mass (g/mol) = Density (g/L) x Molar Volume (L/mol)
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Objectives
Define the terms
Calculate the percent composition of a substance from its chemical formula or experimental data
Derive the empirical formula and the molecular formula of a compound from experimental data
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Terms to Know
Percent composition – relative amounts of each element in a compound
Empirical formula – lowest whole- number ratio of the atoms of an element in a compound
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An 8.20 g piece of magnesium combines completely with 5.40 g of oxygen to form a compound. What is the percent composition of this compound?
1. Calculate the total mass
2. Divide each given by the total mass and then multiply by 100%
3. Check your answer: The percentages should total 100%
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Answer
The total mass is 8.20 g + 5.40 g = 13.60 g
Divide 8.2 g by 13.6 g and then multiply by 100% = 60.29412 = 60.3%
Divide 5.4 g by 13.6 g and then multiply by 100% = 39.70588 = 39.7%
Check your answer: 60.3% + 39.7% = 100%
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Calculate the percent composition of propane (C3H8)
1. List the elements2. Count the atoms3. Multiply the number of atoms of the element by the atomic mass of the element (atomic mass is on the periodic table)4. Express each element as a percentage of the total molar mass5. Check your answer
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Answer
Total molar mass = 44.11 g/mol
36.03 g C = 81.68%
8.08 g H = 18.32%
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Calculate the mass of carbon in 82.0 g of propane (C3H8)
1. Calculate the percent composition using the formula (See previous problem)
2. Determine 81.68% of 82.0 g
Move decimal two places to the
left (.8168 x 82 g)
3. Answer = 66.98 g
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Calculating Empirical Formulas
Microscopic – atoms
Macroscopic – moles of atoms
Lowest whole-number ratio may not be the same as the compound formula
Example: The empirical formula of hydrogen peroxide (H2O2) is HO
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Empirical Formulas
The first step is to find the mole-to-mole ratio of the elements in the compoundIf the numbers are both whole numbers, these will be the subscripts of the elements in the formulaIf the whole numbers are identical, substitute the number 1
Example: C2H2 and C8H8 have an empirical formula of CHIf either or both numbers are not whole numbers, numbers in the ratio must be multiplied by the same number to yield whole number subscripts
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What is the empirical formula of a compound that is 25.9% nitrogen and 74.1% oxygen?
1. Assume 100 g of the compound, so that there are 25.9 g N and 74.1 g O
2. Convert to mole-to-mole ratio:
Divide each by mass of one mole
25.9 g divided by 14.01 g = 1.85 mol N
74.1 g divided by 16.00 g = 4.63 mol O
3. Divide both molar quantities by the smaller number of moles
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4. 1.85/1.85 = 1 mol N 4.63/1.85 = 2.5 mol O
5. Multiply by a number that converts each to a whole number (In this case, the number is 2 because 2 x 2.5 = 5, which is the smallest whole number )2 x 1 mol N = 22 x 2.5 mol O = 5Answer: The empirical formula is N2O5
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Determine the Empirical Formulas
1. H2O2
2. CO2
3. N2H4
4. C6H12O6
5. What is the empirical formula of a compound that is 3.7% H, 44.4% C, and 51.9% N?
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Answers
Compound Empirical Formula
1. H2O2 HO
2. CO2 CO2
3. N2H4 NH2
4. C6H12O6 CH2O
5. HCN
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Calculating Molecular FormulasThe molar mass of a compound is a simple whole-number multiple of the molar mass of the empirical formula
The molecular formula may or may not be the same as the empirical formula
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Calculate the molecular formula of the compound whose molar mass is 60.12 g and empirical formula is CH4N.
1. Using the empirical formula, calculate the empirical formula mass (efm)
(Use the same procedure used to calculate molar mass.)2. Divide the known molar mass by the efm 3. Multiply the formula subscripts by this value to get the molecular formula
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Answer
Molar mass (efm) is 30.06 g
60.12 g divided by 30.06 g = 2
Answer: C2H8N2