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Unit 7
Quantity in Chemistry
Unit 7 Learning Goals
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Define & differentiate between atom, molecule and formula unit
Define & differentiate between atomic mass and formula mass
Determine the molar mass when given symbol or formula
Define mole and Avogadro’s number
Convert moles to mass and mass to moles
Convert particles to moles and moles to particles Define molar volume and convert liters to moles & moles to liters
Determine % composition when given formula or name for compound
Determine empirical formula when given % composition
Define & differentiate between molecular formula and empirical formula
Determine empirical formula when given molecular formula
Determine molecular formula when given empirical formula and formula mass Key Terms – Part One
Microscopic View of Matter
• Atom • Atomic mass
• Atomic element
• Diatomic element
• Molecule
• Formula unit
• Formula mass
Macroscopic View of Matter
• Mole
• Molar mass
• Molar volume
• Avogadro’s Number
Learning Goals
• Define & differentiate between atom, molecule and formula unit
• Identify the elements that are diatomic and describe difference between diatomic and atomic elements
• Define & differentiate between atomic mass and formula mass
Atom and Atomic Mass
• Atom is the simplest form of an element that represents that element
• Each element has a unique symbol (ex: Fe)
• Atomic mass is the weighted average of the mass numbers of all the naturally‐occurring isotopes of that element
Atomic Element and Diatomic Element
• Atomic element is an element that exists in nature as a single atom
– Fe, He, S, U, Cu...
• Diatomic element is an element that exists in nature as two atoms covalently bonded together
– There are only 7 (GEN – U – INE)
– H2, N2, O2, F2, Cl2, Br2, I2
Molecule and Formula Unit: Calculating Formula Mass
• One unit of a covalent compound or diatomic element is called a molecule of that substance (ex: H2O; Br2)
• One unit of an ionic compound is called a formula unit
of that compound (ex: NaCl)
• The formula mass of a molecule or formula unit is the sum of all the atomic masses of the elements of which it is comprised
• One molecule of H2O = 2xH + 1xO = (2x1) + 16 = 18amu
• One formula unit of NaCl = 23 + 35.5 = 58.5amu
• One molecule of Br2 = 80 + 80 = 160amu
diatomic element
element
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Ionic compound
+
Learning Goals
• Determine the molar mass when given symbol or formula
• Define mole and Avogadro’s number
• Convert moles to mass and mass to moles
• Convert particles to moles and moles to particles • Define molar volume and convert liters to moles & moles to liters
Mole
• Mole is the “chemists’ serving” used to count only the extremely small particles that make up matter
• The particles that make up a mole could be:
– atoms (Fe)
– molecules (H2O or O2)
– formula units (NaCl) Avogadro’s Number
• Number of particles in one mole
• It is exactly 6.02 x 1023
• The number of atoms in one mole of atoms
• The number of molecules in one mole of molecules
• The number of formula units in one mole of formula units
From amu to grams: what you need to know
1 amu = mass of 1 proton
1 amu = 1.67x10‐24 grams
atomic mass (aka gram atomic mass) = mass of 1 mole of an element in grams
1 mole = 6.02x1023 particles
Atomic mass = (mass in amu) x (# of atoms in 1 mole) x (mass of 1amu in grams)
Fe = .
.
_
= 56g/mole
How to do calculation
From amu to grams: what you need to know
1 amu = mass of 1 proton
1 amu = 1.67x10‐24 grams
molar mass = mass of 1 mole of any substance in grams
1 mole = 6.02x1023 particles
Molar mass = (mass in amu) x (# of particles in 1 mole) x (mass of 1amu in grams)
H2O = .
.
_
= 18g/mole
How to do calculation
Molar Mass and Molar Volume
• Molar mass is the mass, in grams, of one mole of any substance – 1 mole Fe = 56 g
– 1 mole H20 = 18 g
– 1 mole NaCl = 58.5 g
• Molar volume is the volume, in liters, of one mole of any gas at STP (std. temp & pressure)
– 1 mole He gas = 22.4 L
– 1 mole N2 gas = 22.4 L
– 1 mole CO2 gas = 22.4 L
Molar Mass of Selected Substances
Substance
Type of Substance
Atomic or Molar Mass Formula Mass of (mass of one one particle
mole)
Calcium (Ca)
Atomic element
40.1 amu
40.1 grams
Carbon (C0
Atomic element
12.0 amu
12.0 grams
Hydrogen (H2)
Diatomic element 2.0 amu
2.0 grams
Hydrogen peroxide (H2O2)
Covalent
compound
34.0 amu
34.0 grams
Sodium chloride
(NaCl)
Ionic compound
58.5 amu
58.5 grams
Calcium chloride (CaCl2)
Ionic compound
111.0 amu
111.0 grams
Mass of One Mole of Selected Substances
Substance
Representative Particle
Number of Particles in 1 Mole
Mass of 1 mole
Neon (Ne)
Atom
6.02 x 1023
20.2 grams
Zinc (Zn)
Atom
6.02 x 1023
65.4 grams
Oxygen (O2)
Molecule 6.02 x 1023
32.0 grams
Water (H2O)
Molecule
6.02 x 1023
18.0 grams
Magnesium oxide Formula unit
(MgO)
6.02 x 1023
40.3 grams
Calcium fluoride (CaF2)
6.02 x 1023
78.1 grams
Formula unit
Multiplying and Dividing with Exponents
• Positive vs negative exponents
– 3.2 x 102 = 320
– 3.2 x 10‐2 = 0.0320
• (6.02 x 1023) x 3 = (18.06 x 1023) = 1.806 x 1024
• (6.02 x 1023) ÷ 4 = (1.505 x 1023)
• (8.74 x 1023) ÷ (6.02 x 1023) = 1.45
• (7.9 x 1024) ÷ (6.02 x 1023) = (7.9 x 1024 ) ÷ (0.602 x 1024 ) = 13.2 How do you do mole conversions using ratios?
1) You will create a ratio that reflects the information and “ask” in the problem.
How many moles in 150g of NaCl? 2) Create an equivalent ratio using either molar mass, molar volume or Avogadro’s number
3) Solve for X
X = 2.56 mol NaCl
EXAMPLES: Solving Using Ratios
1) How many moles in 150g of NaCl?
.
X = 2.56 mol
2) How many molecules are in 5.5 moles of CO2?
. .
X =3.31 x1024 molecules
3) How many moles are in 50L of N2?
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X = 2.23 moles
4) How many grams are in 3.8 moles of Al2(CO3)3?

. X =889.2 grams
More Practice • Determine the following: – molar mass Mg(OH)2
– molar volume CO2 (gas)
– molecules of CH4 in 5 moles
– moles of C6H12O6 in 360 grams
– moles of N2 gas in 11.2 liters
ACTIVITY: % Composition of M&Ms
• Count the total number of M&Ms
• Determine the mass of total of all M&Ms
• Count the total number of each color in sample
• Determine the mass of each color (all)
• Calculate the % by number of each color
• Calculate the % by mass of each color
Analysis: Is the % by number the same as the % by mass? Why or why not?
Key Terms – Round 2
• Percent composition
– Of formula
– By mass
– By volume
• Hydrated compounds
• Empirical Formula
• Molecular Formula
Learning Goals
• Define percent composition of elements in compound based on molar mass
• Use percent composition to determine actual mass of an element in a known quantity of a compound
• Determine percent composition when given actual amount of each element in compound
• Determine the percent water in a hydrated compound
Percent Composition
The mass of each element in a compound compared to the entire mass of the compound and multiplied by 100
Mass Element
Mass Compound
X 100
Calculating:
Percent Composition by mass
Calculate: Percentage composition of Fe in Fe2O3
Mass of Fe2
Mass of Fe2O3
X 100
(56g) x 2
[(56g) x 2] + [(16g) x 3]
X 100
Answer: Percentage composition of Fe = 70%
Calculating:
Percent Composition by mass
Calculate: Percentage composition of O in Fe2O3
Mass of O3
Mass of Fe2O3
X 100
(16g) x 3
[(56g) x 2] + [(16g) x 3]
X 100
Answer: Percentage composition of O = 30%
Calculating: Mass of element in compound Given: % composition and total mass of compound
Calculate: Grams of O and Fe in 15.2 grams of Fe2O3
30% Oxygen X 15.2 grams of Fe2O3 = 4.56 grams O
70% Iron X 15.2 grams of Fe2O3 = 10.64 grams Fe
Calculating: Mass of element in compound Given: mass of each element in a compound
Calculate: 5.7g of Ca combine completely with 2.1g of Cl to form a compound. What is the percent composition of the elements in this compound?
Total mass of compound = 5.7g + 2.1g = 7.8g
Mass of Ca ÷ Mass of Compound = (5.7 ÷ 7.8) x 100 = 73%
Mass of Cl ÷ Mass of Compound (2.1 ÷ 7.8) x 100 = 27%
Hydrated Compounds
• Ionic compounds with a specific amount of water • Examples:
MgSO4 • 5H2O = Magnesium sulfate pentahydrate
KNO3 • 3H2O = Potassium nitrate trihydrate
Cu3PO4 • 7H2O = Copper (I) phosphate heptahydrate
Calculating: Percent of Water in Hydrated Compounds
• MgSO4 • 5H2O Magnesium sulfate pentahydrate
Calculate: Percentage composition of H2O in MgSO4 • 5H2O
Mass of 5H2O
Mass of MgSO4 • 5H2O
(18g) x 5
[(24g) x 1] + [(32g) x 1] + [(16g) x 4] + [(18g) x 5]
Answer: Percentage composition of H2O = 43%
x 100 Learning Goals
 Determine empirical formula when given % composition
 Define and identify empirical formulas
What is a chemical formula?
• A qualitative description of what is in a compound
CH2O
1 mol C 2 mol H 1 mol O
• A quantitative description of how much of a component is in a compound
• When compound formulas are described as mole ratios of atoms or ions they can be connected to mass easier
C6H12O6
6 mol C 12 mol H 6 mol O
• Subscripts represent mole ratios of atoms or ions
Empirical Formula
• An empirical formula is the simplest formula for a compound
• It gives the simplest whole‐number ratio
of the atoms of the elements
• The empirical formula is based on the percent composition or atom to atom ratio of the elements in that compound
• Helps define the mass ratios of atoms or ions, and hence identify the presence of a substance
Calculation of Empirical Formula from Data
• Percent to mass
Change the % unit to gram by replacing it (we presume 100 grams of sample)
• Mass to Mole
Convert each gram amount to the mole amount (mole bridge)
• Divide by small
Divide each mole amount by the smallest one obtained
• Multiply ‘til whole
If the previous step did not produce whole numbers (± 0.1) find a multiplier that can convert all numbers to whole, or “near” (± 0.1) whole numbers
Determining Empirical Formula
Ex: A compound is 31.9% potassium, 28.9% chlorine and 39.2% oxygen. what is empirical formula?
– 31.9g ÷ 39g/mole K = .82 mol K ÷ .81 = 1
– 28.9g ÷ 35.5g/mole Cl = .81 mol Cl ÷ .81 = 1
– 39.2g ÷ 16g/mole O = 2.45 mol O ÷ .81 = 3
Atom to atom ratio is 1:1:3
Empirical formula is KClO3
Determining Empirical Formula
Ex: A compound is 50.05% sulfur and 49.95% oxygen. What is empirical formula?
_____g ÷ _____g/mole S = _____ mol S ÷ _____ = ____
_____g ÷ _____g/mole S = _____ mol S ÷ _____ = ____
Atom to atom ratio is _________
Empirical formula is ____________
http://firstyear.chem.usyd.edu.au/calculators/empirical_formula.shtml
Determining Empirical Formula
Ex: A compound is 64.8% carbon, 13.62% hydrogen and 21.58% oxygen. What is empirical formula?
_____g ÷ _____g/mole S = _____ mol S ÷ _____ = ____
_____g ÷ _____g/mole S = _____ mol S ÷ _____ = ____
_____g ÷ _____g/mole S = _____ mol S ÷ _____ = ____
Atom to atom ratio is _________
Empirical formula is ____________
Determining Empirical Formula
Ex: A compound is 31.42% sulfur, 31.35% oxygen and 37.23% fluorine. What is empirical formula?
Atom to atom ratio is _________
Empirical formula is ____________
Determining Empirical Formula
Ex: A compound is found to contain 23.3% magnesium, 30.7% sulfur and 46.0% oxygen. What is the empirical formula of this compound?
_____g ÷ _____g/mole Mg = ____ mol Mg ÷ ____ = ____
_____g ÷ _____g/mole S = _____ mol S ÷ _____ = ____
_____g ÷ _____g/mole O = _____ mol O ÷ _____ = ____
Atom to atom ratio is _________
Empirical formula is ____________
Try This:
NutraSweet is 57.14% C, 6.16% H, 9.52% N, and 27.18% O. Calculate the empirical formula of NutraSweet.
_____g ÷ ___g/mole C = ____ mol C ÷ ____ =___ X ___ =___
_____g ÷ ___g/mole H = ____ mol H ÷ ____ =___ X ___ =___
_____g ÷ ___g/mole N = ____ mol N ÷ ____ =___ X ___ =___
_____g ÷ ___g/mole O = ____ mol O ÷ ____ =___ X ___ =___
http://firstyear.chem.usyd.edu.au/calculators/empirical_formula.shtml
Learning Goals
• Define and differentiate between empirical formula and molecular formula
• Determine empirical formula when given molecular formula
• Determine molecular formula when given empirical formula and molar mass of molecular formula.
Molecular Formula
• A molecular formula is the actual ratio of atoms in a compound
• It is either the same as, or a multiple of, the empirical formula
For ionic compounds, it is ALWAYS the same as the empirical formula
Examples:
lead (II) nitrate‐ Pb(NO3)2
carbon dioxide‐ CO2
sodium chloride – NaCl
glucose – C6H12O6
Molecular Formula to Empirical Formula Problems
Ex: Glucose is C6H12O6 what is empirical formula?
– atom to atom ratio is 1:2:1 – empirical formula is CH2O
Comparing Molecular and Empirical Formulas
Compound
Molecular Formula
Empirical Formula
Glucose
C6H12O6
CH2O
Ibuprofen C13H18O2
Caffeine
C8H10N4O2
Ethylene glycol C2H6O2
Acetaminophen
C8H9NO2
Comparing Molecular and Empirical Formulas
Compound
Molecular Formula
acetylene
C2H2
benzene
C 6 H6
formaldehyde
CH2O
acetic acid
C2H4O2
glucose
C6H12O6
Empirical Formula
Calculating Molecular Formula from Empirical Formula Data
• Determine a common factor by comparing the molar mass of the unknown compound with the formula mass of the empirical formula (efm)
• Apply that factor to the empirical formula as a multiplier
• The result is the molecular formula
Calculating Molecular Formula from Empirical Formula Data
• Empirical Mass
Determine the empirical formula mass
• Divide for whole
Divide the molar mass by the empirical mass
• Multiply for mole
Multiply the whole number by each atom in the compound to get molecular formula
Calculating Molecular Formula
An empirical formula is CH4 and the molar mass is 48 g/mol. What is molecular formula?
1.
Determine the efm (empirical formula mass)
2.
Divide the molar mass by the (efm) empirical formula mass to achieve a whole #
3.
Use this whole # as a multiplier for every type of atom in the empirical formula to get the molecular formula
Calculating Molecular Formula
What is the molecular formula of a compound that has a molar mass of 118.0 g/mol and an empirical formula of C3H7O
1.
Determine the efm (empirical formula mass)
2.
Divide the molar mass by the (efm) empirical formula mass to achieve a whole #
3.
Use this whole # as a multiplier for every type of atom in the empirical formula to get the molecular formula
Calculating Molecular Formula
What is the molecular formula of a compound that has a molar mass of 60.0 g/mol and an empirical formula of CH4N.
1.
Determine the efm (empirical formula mass)
2.
Divide the molar mass by the (efm) empirical formula mass to achieve a whole #
3.
Use this whole # as a multiplier for every type of atom in the empirical formula to get the molecular formula
Empirical/Molecular Formula Problems
Ex: An empirical formula is CH4 and the formula mass is 48amu. What is molecular formula?
– 48amu ÷ mass of CH4 = 48 ÷ 16 = 3 (this is multiplier)
– molecular formula is C3H12 (3 x CH4 )
Ex: A compound is 63% Mn, 37% O. (Hint: presume 100g and change % to grams)
– 63g ÷ 55g/mole Mn = 1.1 mol Mn
– 37g ÷ 16g/mole O = 2.3 mol O
– Atom to atom ratio is 1:2
– Empirical formula is MnO2
Empirical/Molecular Formula Problems
Ex: The molecular formula for hydrogen peroxide is H2O2. What is the empirical formula?
– Divide formula by least common denominator
– empirical formula is HO
(H2O2 ÷ 2 )
Ex: A compound contains 56g Nitrogen and 160g Oxygen. What is the empirical formula? What is the molecular formula?
– 56g ÷ 14g/mole N = 4 mol N
– 160g ÷ 16g/mole O = 10 mol O
– Atom to atom ratio is 2:5
– Empirical formula is N2O5
– Molecular formula is N4O10