CHAPTER 8: CHEMICAL COMPOSITION

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CHAPTER 8: CHEMICAL COMPOSITION

Active Learning: 1-4, 6-8, 12, 18-25;

End-of-Chapter Problems: 3-4, 9-82, 84-85, 87-92, 94-104, 107-109, 111, 113, 119, 125-126

8.2 ATOMIC MASSES: COUNTING ATOMS BY WEIGHING

Atoms are too small to weigh directly

– e.g. one carbon atom has a mass of 1.99×10-23_{g—too inconvenient an amount to use!}

→

need more convenient unit for an atom’s mass

→

atomic mass unit (amu)

Carbon-12 was chosen as the reference and given a mass value of 12 amu

→

1 amu = 1/12th the mass of a carbon-12 atom

→

Masses for all other elements are measured relative to mass of a carbon-12 atom

Weighted Average Atomic Mass of an Element

– Why is carbon’s mass on Periodic Table 12.01 amu, NOT 12.00 amu?!

– Atomic masses reported on the Periodic Table are weighted averages of the masses of all the naturally occurring isotopes for each element based on their percent natural

abundance—i.e., the percentage of existing atoms that are a specific naturally occurring

isotopes.

Ex. 1 If 98.892% of carbon exists as carbon-12, which has a mass of 12.00000, while 1.108% exists as carbon-13, which has a mass of 13.00335, calculate the average atomic mass for carbon.

average atomic mass = (0.98892)(12.00000 amu) + (0.01108)(13.00335 amu)

Ex. 2 The atomic masses of the three naturally occurring isotopes or argon, Ar-36 (0.3365%), Ar-38 (0.0632%) and Ar-40 (99.6003%), are 35.96754552 amu, 37.9627325 amu, and 39.9623837 amu, respectively. Calculate the average atomic mass for argon.

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8.3 THE MOLE

Atomic Masses and Molar Masses

If 98.93% of C atoms are carbon-12 and 1.07% of C atoms are carbon-13, then the (weighted average) atomic mass of carbon is calculated as follows:

(12.000000 amu)(0.9893) + (13.003354 amu)(0.0107) = 12.01 amu So how many carbon atoms are present in 12.01 grams of carbon?

This number was determined experimentally to be 6.022×1023.

– It was named Avogadro’s number, to honor Italian scientist Amedeo Avogadro who first proposed in 1811 that the volume of a gas depended only on the number (not type) of atoms or molecules present at a given temperature and pressure

AVOGADRO'S NUMBER (NA)= 6.022×1023 (4 sig figs)

How big is this?

• If 6.022×1023 hydrogen atoms were laid side by side, the total length would encircle the

Earth about a million times.

• The mass of 6.022×1023 Olympic shotput balls is about equal to the mass of the Earth.

• The volume of 6.022×1023 softballs is about equal to the volume of the Earth.

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THE MOLE CONCEPT

1 mole (abbreviated mol) = 6.022×1023_items

Similar to: 1 dozen = 12 items

1 dozen doughnuts = 12 doughnuts 1 mole of doughnuts = 6.022×1023 doughnuts

MOLE CALCULATIONS I

a. How many eggs are in 3 dozen eggs? ____________ b. How many eggs are in 3.00 moles of eggs?

c. How many moles of C atoms are present in a sample of 1.25×1024 C atoms?

8.5 MOLAR MASS

Thus, the mass of 1 mole of C atoms is 12.01 g (or 12.01 g/mol)

→

1 mole (6.022×1023) is the amount of atoms of any element that has a mass in grams equal to the mass of ONE atom in amu.

→

The atomic masses reported for each element in the Periodic Table give the average

atomic mass in amu and the molar mass in g/mol.

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Molar mass (MM): Mass in grams of 1 mole of any element or compound

– To obtain the molar mass of a compound, multiply the molar mass of each element by the number of each present, then add them all up.

Example: Determine the molar mass of each of the following compounds: a. O2: 2 (molar mass of O) = 2 (16.00 g/mol) = 32.00 g/mol

b. NaCl: c. CO2:

d. H2SO4: e. Ca3(PO4)2:

Do not worry about rounding molar mass to the correct number of sig figs! Molar

mass will rarely be the measurement that limits the number of sig figs for a calculation.

CONVERSIONS AMONG MASS, NUMBER OF MOLES, AND NUMBER OF PARTICLES

Use the dimensional/unit analysis method: 1. Write the units of the final answer.

2. Write the given information related to the answer.

3. Determine the unit factors (Avogadro’s #, molar masses) necessary to solve the problem using the given information

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Ex. 2: How many moles of neon are in 25.0 g of neon?

Ex. 3: How many neon atoms are in 25.0 g of neon?

Ex. 4: How many moles of H2O are in 50.0 g of H2O?

Ex. 5: How many H2O molecules are in 50.0 g of H2O?

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MOLAR VOLUME: The volume occupied by 1 mole of any gas

Avogadro's Law: At the same temperature and pressure, equal volumes of gases contain the same number of molecules.

Standard temperature and pressure

(STP): T=0˚C and P=1.00 atm At STP,

1 mole

of gas occupies

22.4 L

! 3 sig figs Write 2 unit factors using this information:

Mole Calculations with Molar Volume:

Ex. 1 What mass of CO2 occupies a volume of 25.0 L at STP?

Ex. 2 What is the volume occupied by 10.0 g of propane gas (C3H8) at STP?

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8.6 PERCENT COMPOSITION OF COMPOUNDS

Mass Percent Composition (or Mass Percent): The mass of each element in a compound

divided by the mass of the entire compound.

100% Compound of Mass Total Element of Mass = Element of Percent Mass ×

Steps to determine percentage composition:

1. Calculate the total mass of each individual element in the compound

2. Add up all the masses of each element to get the total mass of the compound 3. Divide the mass of each individual element with the total mass of compound Ex. 1 What is the percent composition by mass of each element in H2O?

____________% H ____________% O Ex. 2 What is the percent composition by mass of each element in K2SO4?

____________% K ____________% S ____________% O

Ex. 3 What is the percent composition by mass of each element in Al2(CO3)3?

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Ex. 4: Calculate the mass of iron in 50.0 g of rust, iron(III) oxide.

Ex. 5: Calculate the mass of silver present in 125 g of silver carbonate.

Ex. 6 What mass of nitrogen is present in 75.0 g of ammonium nitrate?

8.7 FORMULAS OF COMPOUNDS

Empirical Formula: Simplest whole-number ratio of atoms in a compound

– where the term “empirical” means “derived from experiment”

Molecular Formula: Chemical formula of a compound that expresses the actual number of

atoms present in one molecule.

– The molecular formula will either be exactly the same or some multiple of the empirical formula!

Fill in the table below:

glucose, C6H12O6 caffeine, C6H10N4O2 acrylonitrile, C3H3N

Empirical Formula Molecular Formula

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8.8 CALCULATION OF EMPIRICAL FORMULAS

Guidelines for Determining the Empirical Formula of a Compound

1. Find the # of moles of each element in the compound.

2. Divide each # of moles of each element by smallest # of moles to get ratio of atoms. 3. Get a whole number ratio for all atoms in the compound:

– If within 0.1 of a whole number, round to that whole number – If any ratio ends close to 0.5

_→

multiply ALL subscripts by 2

– If any ratio ends close to 0.33 or 0.66

→

multiply ALL subscripts by 3

Ex. 1: Determine the empirical formula for nickel oxide if a sample of nickel oxide consists of 17.74 g of nickel and 7.26 g of oxygen?

empirical formula: ______________ name: __________________________ Ex. 2: Ascorbic acid, known more commonly as Vitamin C, is a hydrocarbon derivative, a

compound that consists of carbon, hydrogen, and oxygen. If a 25.00 g sample of ascorbic acid contains 10.23 g of carbon and 1.14 g of hydrogen, determine the empirical formula for ascorbic acid.

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8.9 CALCULATION OF MOLECULAR FORMULAS Determining the Molecular Formula of a Compound

1. For the Molecular Formula, you will be given the molar mass of the compound, and you need to calculate the molar mass of the empirical formula.

2. Divide the molar mass of the compound by the molar mass of the empirical formula to get the factor by which to multiply each subscript in the empirical formula.

Ex. 3: If the ascorbic acid in Ex. 2 above has a molar mass of 176.1 g/mol, determine its molecular formula.

Empirical and Molecular Formulas from Percent Composition:

1. Assume 100.0 g of the compound is present, so change percent units to grams. 2. Follow the same steps for determining empirical formula and molecular formula.

Ex. 4: Quinine is used as an antimalarial drug. An analysis of quinine indicates the compound consists of 74.03% carbon, 7.47% hydrogen, 8.64% nitrogen, and 9.86% oxygen. If quinine's molar mass is 325 g/mol, determine the empirical and molecular formulas for quinine.

Empirical Formula of Quinine: __________________ Molecular Formula of Quinine: __________________