Measurements in Chemistry - understanding and using the mole

General Chemistry Unit 6 Note Pocket Stoichiometry

Name:__________________ Date: _________ Per: ____

Measurements in Chemistry - understanding and using the mole

Chemical measurements

In chemistry we make some measurements by counting and other measurements by weighing, determining volume, etc. Which technique we employ is determined, in large part, by our purpose. It is also necessary, when determining which technique to use, to consider what type of measurement is easiest to make or even feasible. Consider the chemical reaction below:

C1

(g) + 2KI(aq) —> 2KCl(aq) + I

(s)

You may at first think about the fact that one molecule of chlorine reacts with two formula units of potassium iodide to produce two formula units of potassium chloride and one molecule of iodine.

While this is true, it’s impossible to actually study this reaction at the atomic level in a chemistry laboratory; we cannot count out just one or two atoms or molecules of anything! We have to use another way to measure out the necessary quantities of chlorine and potassium iodide.

The mole

A mole is a quantity of something just like a dozen (12), gross (144), or ream (500). We use these more familiar quantities to count out large quantities because it’s more convenient. We can convert from the individual item to the quantity units by using simple conversion factors. Try the problems below using factor labeling:

5 dozen eggs = ______________ eggs

1152 pencils = _______________ gross of pencils

5000 pieces of paper = ______________reams of paper

A mole relates the number of atoms or molecules of a substance to the mass in grams. The conversion factor used is based on the mass of each element in atomic mass units and how that relates to the mass of the element in grams.

• mole = the number of atoms of an element that would be found in, say, 12.01 grams of carbon-12 or 14.01 grams of nitrogen-14 (the molar mass)

• mole = ______________________ atoms (this number is called “________________ number”)

• mole = __________ L of any gas at STP (standard temperature and pressure: 273 K and 1.0 atm) This means that we can use the atomic masses on the periodic table to convert instantly to grams.

The mass of one mole of atoms is called the molar mass. Use the periodic table to determine the following:

one atom of Ar _____ amu

one mole of Ar atoms ____ g = molar mass of Ar one atom of Na = __________ amu

one mole of Na atoms ____ g = molar mass of Na

one atom of Pb _____ amu

one mole of Pb atoms ______ g molar mass of Pb one atom of B ______ amu

one mole of B atoms ______ g = molar mass of B

Simple conversions using mole equivalencies We use mole equivalencies to convert:

• moles to ________ 1 mole of element = __________________________

• moles to the number of ________ 1 mole of element = __________________________

• moles of gas to _______________ 1 mole of gas = _____________________________

For example:

1.50 mol Na = ______ Na 1.50 mol Na x 23.0 g Na = 34.50 g Na = 34.5 g Na (sig figs!) 1 mol Na

Do the following conversions using the mole equivalencies above. Use factor labeling and show your work.

2.0 mol Ni = _______ g Ni

_________mol C = 24.02 g C

5 mol He = _______ atoms He

___________mol Ca = 4.75 x 10

atoms Ca

0.115 mol Xe = __________L of Xe

_____________ mol Ne = 67.2 L of Ne

Atomic masses of elements  formula mass of a compound  molar mass of a compound We can work easily with the mass of a compound in the same way we do with elements. Determine the mass of a unit of sodium chloride or calcium fluoride by adding up __________________of the component elements in their proper _______________. The sum is called the

________________

one atom of Na = ___________ amu one atom of Ca = ____________ amu

one atom of Cl = ____________ amu two atoms of F = ____________ amu

formula mass of NaCl = _______amu formula mass of CaF

= ________ amu

The molar mass of a substance is determined the same way, but you are finding the mass of _________________________ instead of the mass of just one _______________ (for a covalent compound) or one _____________________ (for an ionic compound).

molar mass of NaCl = ________ g molar mass of CaF

= ________ g molar mass of H

O _________ g molar mass of K

SO

= _______ g molar mass of SO

_________ g molar mass of Cl

= __________ g More simple conversions using mole equivalencies

2.0 mol S0

= ___________g SO

_____________ mol F

= 1.4 L F

2.44 x 10

molecules H

O = ______________ mol H

O

0.98 mol Fe = ______________ atoms Fe

___________g 0

= 4.00 L 0

5.643 g CO

= ______________ molecules CO

0.500 L NO = ____________ moles NO

Multi-step conversions, using mole equivalencies

Conversions involving moles, like other conversions, can involve more than one step. When this is the

case, it is convenient to use the following “mole map” to help you plan the conversion.

Example problem:

2.3 x 10

molecules H

= _____ L of H

2.3 x 10

molecules H

x 1 mol H

x 22.4 L H

= 6.02 x 10

molecules H

1 mol H

Practice Problems:

1. 34 g BeCl

= __________ formula units BeCl

2. 44 g H

O = __________ molecules H

O

3. 3.7 x 10

molecules I

= ____________ moles I

4. 6.90 L CH

= _____________ g CH

5. 0.034 g NaF = ____________ formula units NaF

6. 5.66 x 10

atoms Ar = ____________ L Ar

7. 9.0 x 10

molecules Br

= __________________ g Br

8. 0.05 L CO = ______________ molecules CO

9. How many atoms of uranium are in 5.00 g of uranium?

10. What is the volume of 2.6 g of nitrogen gas?

11. What is the mass of 1.0 x 10

molecules of sulfur dioxide?

12. What is the mass of Neon in a sign when the gas tubes have a total volume of 166 mL?

STOICHIOMETRY

Stoichiometry :

In many ways, stoichiometry is the backbone of the most practical part of chemistry; it helps us relate actual quantities (measured by mass or volume) of reactants to products in a chemical reaction. It’s a lot like using a chocolate chip cookie recipe to produce a certain number of cookies of a given size. You have to put the ingredients together in the proper proportions to make a cookie that tastes good!

Consider the following balanced equation: 2Na + Cl

 2NaCl

We can look at this equation in two different ways:

1. at the atomic scale:

2. at the real-world, measurable scale:

Dimensional analysis can be used to relate the moles of reactants and products in a chemical reaction to one another. It is essentially a matter of determining the ratio of reactants and products by looking at the coefficients in the balanced chemical equation. In the reaction above, the ratio of sodium to chlorine is 2:1.

 the ratio of sodium to sodium chloride is

 the ratio of chlorine to sodium chloride is

If you’re starting out with 0.650 moles of sodium and you want to know how many moles of chlorine you need to fully react with the sodium, you can use dimensional analysis:

0.650 mol Na x 1 mol Cl

=

2 mol Na

If you want to produce 6.4 moles of sodium chloride, how many moles of chlorine do you need to use?

Because we can literally measure out the moles of sodium by weighing it (one mole would have a mass of 22.99g) and we can measure out the moles of chlorine by volume (each mole of chlorine gas would have a volume of 22.4 L), we can literally put the reactants together in their proper proportions. This makes

stoichiometry an incredibly powerful but very simple tool - one that you must master in order to be successful in chemistry!

When solving a stoichiometry problem, you’ll always be given a quantity of either a product or reactant that you have to relate to a quantity of another product or reactant. To do this, you’ll use the balanced chemical equation as well as the mole equivalencies you’ve been using in mole problems. These equivalencies are:

 1 mole =

 1 mole =

 1 mole =

The equivalencies you’ll use depend on the specific problem, but the logic is always the same:

quantity of given  moles of given  moles of unknown  quantity of unknown Use the “new and improved” mole map shown below to help you map out a problem-solving plan for any stoichiometry problem.

Given the following equation:

CH

(g) + 2O

(g)  CO

(g) + 2H

O (l)

What mass of oxygen will react with 36.0 g of methane (CH

)?

36.0 g CH

x 1 mol CH

x 2 mol O

x 32.00 g O

=

16.04 g CH

1 mol CH

1 mol O

*Let’s turn this calculation into a statement…

When 36.0g of methane is used up in the previous reaction, it will require ___________ g of oxygen.

1. What mass of water would be produced when oxygen reacts with the 36.0 g of methane?

* Sentence explaining the calculation:

When 36.0 g of methane reacts with plenty of oxygen, it will create _______________ g of water.

2. How many liters of carbon dioxide would be produced from the reaction if 44.0 g of oxygen are consumed?

*Sentence explaining the calculation:

When 44.0 g of oxygen are reacted with plenty of methane, ______________ L of carbon dioxide are produced.

3. What volume of methane is needed to produce 26.8 g of water?

moles of substance B moles of

substance A

liters of substance B

particles of substance A grams of

substance A

liters of substance A

particles of substance B

grams of

substance B

Limiting Reactant/Reagent Problems:

 a stoichiometry problem where quantities of reactants are not available in the EXACT molar ratios

 one reactant is “in excess” and the other is the “limiting reactant”

 given mass/volume/moles of more than one reactant

 you find the actual amount of product(s) produced by determining which reactant LIMITS the reaction

Strategy #1 for finding Limiting Reactant:

1. Write the balanced chemical equation.

2. Solve one stoichiometry problem for each reactant (A & B). Solve each for quantity of the same product (ideally the one you are asked about).

3. Compare answers to step 2. Whichever reactant (A or B) produces the LEAST amount of product is the limiting reactant.

4. The smaller amount of product is the amount actually produced.

5. If necessary, calculate amount of excess reactant left over. (Think – how much excess reactant will you need to completely react with the limiting reactant? How much do you have? How much is extra?)

Example Problem:

What is the limiting reactant when 1.5 moles of hydrogen react with 0.5 moles of oxygen to produce water?

How many grams of water are produced?

Strategy #2 for finding Limiting Reactant:

1. Write the balanced chemical equation.

2. Solve a stoichiometry problem starting from a given quantity of reactant (A). Solve for the necessary quantity of the other reactant (B).

3. Is there enough reactant B to use all of reactant A? If so, A is limiting. If not, B is limiting.

4. Solve one more stoichiometry problem starting with limiting reactant (A or B) and solving for desired product (whichever you are curious about).

5. If necessary, find how much excess reactant is left over (how much wasn’t used up in the reaction?).

Example Problem:

When 2.60 g of hydrogen react with 10.20 g of oxygen, what mass of water will be produced? What is the

limiting reactant?

Percent Yield Problems

 the amount of product produced from a reaction based on a calculation is the expected yield

 the amount of product that is really obtained during the reaction (in a lab or industrial setting) is the actual yield

 reasons for differences between actual and expected yield:

1. some reactants didn’t fully react

2. there were side reactions that differ from the reaction on which the expected yield was based 3. some product was lost during recovery or transfer

 it is useful to determine what percent of the expected yield was actually obtained percent yield = actual yield x 100%

expected yield Example Problem:

A reaction occurs between 500.0 g of carbon disulfide and an excess amount of oxygen gas. The mass of sulfur dioxide produced is 768 g. What is the percent yield for this reaction?

Step 1. Write the balanced chemical equation.

Step 2. Determine the expected yield using stoichiometry

Step 3. Compare the actual yield to the expected yield using the percent yield equation.