Chapter 6: Measuring Matter

3.3 Conversion Factors (IMPORTANT)

The same quantity can be measured or expressed in many ways 1 dollar = 4 quarters = 10 dimes = 20 nickels 1 meter = 10 decimeters = 100 centimeters

Whenever 2 measurements are equal, a ratio of these measurements will equal 1

You can always multiply by 1! Ratios of equivalent measurement is called a

conversion factor

3.4-3.7 Dimensional Analysis & Conversions

One of the best methods for solving problems Uses the units (dimensions) that are part of

measurements to help solve the problem. Let’s start with an easy problem, but

SHOWING ALL WORK To determine which unit goes on the bottom

of the ratio, look at what needs canceling. We can create infinite chains of this!

Your school club has sold 600 tickets to a chili supper fund-raising event. You volunteered to make the chili (sucker). You have a very large pot, and a chili recipe that serves 10. The recipe calls for 2 teaspoons of chili powder. How much chili powder do you need for 600 servings of chili?

DON’T do this in your head. The skills to do this will be needed later!

Step 1: Unknown Number of teaspoons of chili powder

Step 2: Identify what is known 600 servings must be made 10 servings = 2 tsp of chili powder

Step 3: Plan a solution Create conversion factors 10 servings = 2 tsp chili powder

10 servings / 2 tsp chili powder 2 tsp chili powder / 10 servings

Change the unit “servings” into unit “teaspoons of chili powder”

Step 4: Do the calculations

That’s a lot of teaspoons! Let’s figure out how many cups that is, that would be

easier. Known: 3 tsp = 1 tbsp 16 tbsp = 1 cup

servings

rChiliPowdetspservings

10

.2600 tsp

rChiliPowde120

10

1200

What is known? 120 tsp chili powder

Plan a solution. Need to convert tsp tbsp cups

Calculation

cupsxx

xx

tbsp

cup

tsp

tbsptsp5.2

48

120

1631

11120

16

1

3

1

1

120

Once you’re done, make sure all units (except for the ones you want to keep) cancel.

Check this also before doing any actual calculations. This will tell you if you’re right.

Special Conversion

Volume is tricky. Show on board. Not as straight-forward as length for conversions.

1 cm3 = 1 mL If you mess up, and your answer is upside-

down, just take the inverse of it!

6.1 Measuring Matter

One of the most common ways to measure matter is to count it

You can weigh it Measure by volume Some by multiple ways

You can buy nails by mass or count

Some units used in measuring refer to a specific number of items A pair A dozen A baker’s dozen

Some items, like fruit, are commonly measured and purchased in multiple ways. Fruit stand, 5 apples for $2.00 (count) Grocery store, $1.00 per pound (weight) Orchard, $9.00 per bushel (volume)

Each of these is a way to measure apples. We can further extend this to apply to a dozen apples.

1 dozen apples = 12 apples (count) Using average sized apples, 1 dozen apples

= 2 kg apples Also using average sized apples, 1 dozen

apples = 0.20 bushel apples So what we have is a series of conversions

that can be used to calculate number, mass or volume

Unit Conversions!

What is the mass of 90 average-sized apples?

1. What is unknown? Mass of apples

2. What is known? 90 apples, 1 dozen apples = 2 kg, 1 dozen = 12

3. Conversion is Number apples mass Number of apples dozens of apples mass

of apples

sdozenApple

kgApples

apples

sdozenAppleapples

1

0.2

12

190

= 15 kg apples

The Mole More than a reality canceled TV show The mole is nothing more than a certain amount of anything Often used to count atoms, since they are so small, scales like a

“dozen” were impractical A mole represents 6.02x1023 of ANYTHING. A mole of hydrogen molecules is 6.02x1023 hydrogen molecules A mole of cars is 6.02x1023 cars This is a REALLY big number. A mole of eggs would fill the world’s oceans 30 million times

over. It would take 10 billion chickens laying 10 eggs per day more

than 10 billion years to lay a mole of eggs 1 mole of seconds would be 1.9x1016 years, which is

19,000,000,000,000,000 (compared to age of universe which is 20,000,000,000 years)

6.02x1023 is called Avagadro’s number, in honor of Amedeo Avogadro di Quarenga

When dealing with chemistry units, we often have to think about what we’re dealing with

A mole of oxygen is usually referring to 6.02x1023 oxygen molecules, not atoms

A mole of iron is 6.02x1023 atoms of iron A mole of salt is 6.02x1023 formula units of NaCl A mole of water is 6.02x1023 molecules of water Table 6.1 on page 145 details more of this

Some problem solving with this

Handled just like the apple example with a dozen How many moles of magnesium are 3.01x1022 atoms

of magnesium? We now have the conversion 1 mole (mol)

magnesium = 6.02x1023 atoms of magnesium

molMgxatomsMgx

molMgatomsMg 223

22

1051002.6

13.01x10

Note, Avogadro's number is a conversion, so an infinite number of significant figures

6.3-6.4 The Gram Formula Mass, Gram Atomic Mass and Molar Mass This sounds worse than it actually is The gram atomic mass (gam) is the atomic

mass of an element expressed in grams Atomic mass found on periodic table

The gram atomic mass is the mass of one mole of atoms of any element

For single elements, the gram atomic mass = atomic mass.

Mass of mole of a compound

First you must find the formula for the compound

Let’s use sulfur trioxide as an example SO3

One molecule of sulfur trioxide is from one atom of sulfur and three atoms of oxygen.

So one mole of sulfur trioxide contains one mole of sulfur and three moles of oxygen

To find the mass, must find mass of each part, then add together.

Gram atomic mass of sulfur = 32.1 g

Gram atomic mass of oxygen = 16.0 g But there’s three moles of oxygen, so must multiply by 3! So total oxygen contribution is: 16.0x3 =48.0 g

Add them together: 32.1 + 48.0 = 80.1 g That is the mass of one mole of sulfur trioxide. Because we’re dealing with a molecule now, this is

not the gram atomic mass, but the gram molecular mass (gmm).

How about ionic compounds?

Can’t find the gram atomic mass, because not atoms.

Can’t find the gram molecular mass, because not molecules.

They use gram formula mass (gfm) Besides the name, calculated JUST like the gram

molecular mass. So gram formula mass of NaCl is equal to the sum

of the gam of sodium plus the gam of chlorine. Generically, all of these can be called the molar

mass. Not as specific, but still works, and I’ll use it.

6.5 Mole Mass Conversions

We can use the molar mass of a substance to convert grams of a substance into moles.

We can do this since molar mass = gram/mole

We can also use a similar practice to turn moles grams. Will still be using molar mass of the substance.

A mol A massmolar

A mol 1A grams

B grams B mol 1

B massmolar B mol

6.6 The Volume of a Mole of Gas

For many thing, the volume of a mole of a substance is different from most other substances.

For example, a volume for a mole of water is much less than the volume for a mole of sugar.

However, the volume of a mole of gases is MUCH more predictable.

As we will learn later in the year, the volume of a gas varies with the temperature or pressure.

Because of this variation, the volume of a gas is usually measured at a standard temperature and pressure (STP). This is defined as 0°C and 101.3 kPa

(kilopascals) or 1 atmosphere (atm). At STP, 1 mole of ANY gas will have a volume of

22.4 L. Black box is 22.4 L.

The volume, 22.4 L is known as the molar volume of a gas, measured at STP.

This idea of all gases containing the same number of particles in a given volume at any given pressure and temperature is what made Avogadro famous. And is why 6.02x1023 (1 mole of anything) is

named after him. So, 22.4L of any gas at STP contains 6.02x1023

particles of that gas.

Would 22.4 L of one gas (at STP) have the same mass as 22.4 L of another gas (at STP)?

No. Why? Remember from yesterday, molar masses. If the molar mass of the gases is different, so will

the mass of 1 mole of their gas. Only if the molar masses of two gases is the

same will their mass at 22.4 L (at STP) be the same.

Where does this come in handy?

Gives us another conversion At STP, 1 mole (of a gas) = 22.4 L For example,

Determine the number of moles in 33.6 L of He gas at STP. Going to convert L moles

He mol 5.1He L 22.4

He mol 00.1He L 6.33

6.7 Gas Density and the Molar Mass

Density is usually measured in units of g/L. We can use the density of a gas at STP to

determine the molar mass of that gas. When doing so, the gas can be an element,

or a compound.

Example

The density of a gaseous compound of carbon and oxygen is 1.964 g/L at STP. Determine it’s molar mass, and determine if the substance is carbon dioxide or carbon monoxide (important difference here). May seem awkward to start here, but we can do

this. Let’s tackle the steps in solving this.

1. What are we looking for? Molar mass

2. What do we know? Density is 1.964 g/L 1 mole of gas = 22.4 L

3. Outline path Density Molar mass

Which is the same as Density g/mol

mol

g 44.0

mol 1

L 4.22

L 1

g 964.1

•So the molar mass of this gas is 44.0 g/mol

•Almost done, just need to determine if this is CO or CO2

•Calculate molar mass of CO:

•28.0 g/mol

•Calculate molar mass of CO2

•44 g/mol

•This unknown substance is CO2

6.8 Converting Between Units with Moles

We have now examined a mole in terms of particles, mass, and for a gas, volume at STP. 22.4 L = 1 mole Molar mass = 1 mole 6.02x1023 particles = 1 mole All of these have 1 mole in common, so we can

always use the mole to convert from one to the other!

6.9 Percent Composition When we make a new compound in the laboratory,

we need to determine its formula. One of the first steps is to find the relative amounts

of the elements in the compound. These relative amounts are expressed as the

percent composition. This looks at the mass of each element in the compound There are as many percent values in a compound as there

are different elements in the compound. For example, K2Cr2O7 is K = 26.5%, Cr = 35.4% and 0 =

38.1 %. These percents must add up to 100%.

The percent by mass of an element in a compound is the number of grams of the element divided by the grams of the compound, multiplied by 100%

Or,

%100compound of grams

Eelement of gramsEelement of mass %

Example

An 8.20-g piece of Mg combines completely with 5.40 g of oxygen to form a compound. What is the percent composition of this compound?

Since we know the mass of each element, we can add them to find the mass of the compound: 8.20g + 5.40g = 13.60g

Now we divide the mass of each element by the mass of the compound, then multiply by 100%

%10013.6g

g20.8%100

compound of mass

Mg of massMg %

%3.60

%10013.6g

g40.5%100

compound of mass

O of massO %

%7.39

Check your percentages to make sure they add to up 100%...

And we’re done! 60.3% magnesium and 39.7% oxygen by mass.

% Composition of a Known Compound

To do this, we will use the chemical formula of the compound to calculate the molar mass. This gives me the mass of one mole of the compound.

Then for each element, calculate the percent by mass in ONE MOLE of the compound.

The subscripts in the formula are used to calculate the grams of each element in a mole of that compound.

%100compound of massmolar

compound of mol 1in element of gramsmass %

Example

Calculate the percent composition of propane, C3H8

First, calculate the molar mass of propane 3 mol C =

36.0g C 8 mol H =

8.0 g H Molar mass of propane = 44.0 g

Now percent by mass for each element can be calculated, just like before!

%10044.0g

g0.36%100

compound of massmolar

C of massC %

C %8.81

%10044.0g

g0.8%100

compound of massmolar

H of massH %

H %2.18

% Composition with a specific amount of a substance

So far, these amounts have been generic amounts for the compound. Propane in general, etc.

You can use the % composition to calculate the number of grams of an element in a specific amount of a compound.

To do so, you multiply the mass of the compound by the conversion factor based on the % composition.

Example

Using propane from previous example. Calculate mass of carbon in 82.0 g of

propane, C3H8

From the previous example, I know that % composition of propane is 81.8% C and 18.2% H. This means that for every 100 g of propane, you

will have 81.8 g of carbon And for every 100 g of propane, you will have

18.2 g H.

So we convert….

C g1.67HC g 100

C g 81.8 HC 82g

8383

•So 82g of propane, there will be 67.1 g of carbon.

6.10 Calculating Empirical Formulas

Once a new compound has been made in the lab, you can usually determine its % composition experimentally.

Then, from % composition data, you can calculate its empirical formula. The empirical formula gives the lowest whole-

number ratio of the atoms of the elements in the compound

Example A compound may have an empirical formula of CO. This tells me that C and O have a 1:1 ratio However, empirical formula may or may not be the same

as the actual molecular formula. If not the same as the molecular formula, the molecular

formula will be a simple multiple of the empirical formula Dinitrogen tetrahydride has formula of N2H4

But has empirical formula of NH2

How to Calculate Empirical Formulas

This is a little hard. Just giving you a head’s up.

First, you must know the % composition. Let’s give this a shot... We’ll go through step-

by-step. Example:

What is the empirical formula of a compound that is 25.9% nitrogen and 74.1% oxygen?

Because 25.9% nitrogen, and 74.1% oxygen, we can say that in 100 g of the compound There will be 25.9 g

nitrogen There will be 74.1 g

oxygen Need to convert these

masses into moles To do this, I use molar

mass of these elements

N mol 85.1N g 14.0

N mol 1N g 25.9

O mol 63.4O g 16.0

O mol 1O g 74.1

We now have a mole ratio of nitrogen to oxygen of: N1.85O4.63

This doesn’t work. We want whole numbers. To start out, divide both numbers by the

smaller number of moles. This this case, divide both by 1.85

1.85 / 1.85 = 1 mol N 4.63 / 1.85 = 2.5 mol O

Now have a mole ratio of NO2.5

Still doesn’t work, need whole numbers.

Now must multiply both by a number to convert the fraction to a whole number. This takes practice.

2.5 a whole number, what easy multiplication could I use?

2 So…

1 mol N x 2 = 2 mol N 2.5 mol O x 2 = 5 mol O

Gives us a mole ratio of N2O5 If both numbers are whole numbers, this is our

empirical formula

Overview, step-by-step…1. Need % composition2. Determine mass of each element in 100g of

compound Hint: Just trade % for g

3. Find # of moles of each element Use molar mass for this

4. Divide both mole ratios by lowest number of moles

5. Multiply both mole ratios by a number to bring higher mole ratio to a whole number

6.11 Calculating Molecular Formulas

We know how to calculate empirical formulas, but remember, empirical formulas don’t always equal molecular formulas

For example, Ethyne and benzene each have empirical formula of CH Glucose, ethanoic acid and methanal all have empirical

formula of CH2O

For each of these that have the same empirical formula but are different compounds, they will also have different molar masses (gram formula mass)

For each of these with the same empirical but different gfm, the gfm are all simple, whole-number multiples of the efm (empirical formula mass) of the empirical formula. Example:

A molecule’s empirical gfm is 12. It’s real gfm will be 12, 24, 36, 48, etc.

We can determine the molecular formula of the compound IF we know the empirical formula and its gfm.

To do this, divide the gram formula mass by the empirical formula mass.

This gives the whole-number multiple to apply to the empirical formula.

Examples

For each of these, calculate the molecular formulas of the given compounds

1. Gram formula mass of 60 g, empirical formula of CH4N

First need to find efm. Efm = 30 Divide gfm/efm 60/30 = 2 Apply this multiple to the empirical formula Giving us a molecular formula of C2H8N2

2. Gram formula mass of 78 g, empirical formula of NaO

Same steps, first we find the efm. Efm = 39 Gfm/efm = 78/39 = 2 (again) Apply this multiple to the empirical formula Molecular formula is Na2O2

3. Gram formula mass of 181.5 g, empirical formula of C2HCl

What would this be? Figure it out. You will need:

Empirical formula mass Gram formula mass Empirical formula

Molecular formula is C6H3Cl3

If the problem does not give the empirical formula, what would you need?

The percent composition Or, enough information to calculate %

composition. That’s it, done with chapter!