Unit 2

Venus, the 2nd

planet of our solar system

Ch. 3:

Conversions between two substances focus

primarily in the ratio given by the balanced

chemical equation (BCE). The primary method to

go from any unit to any unit of another substance is by converting to moles.

To convert to moles from:

Mass

☼ Divide by the molar mass of the substance, g/mol

Formula units

☼ Divide by Avogadro’s number, 6.02 x 1023

form. units/mol

Volume

☼ You have the option to go from volume to volume of the substances, so

long as both are gases. Otherwise, divide by 22.4L/mol, as long as it is at

STP conditions (1 atm. and 0°C) If not at STP, see unit 3.

Below is the balanced chemical equation for the combustion of propane. Given 12.0g of propane,

find how many grams of CO2 will be produced.

C3H8+ 5O2→4H2O + 3CO2

Starting with 12.0g of propane, you convert this amount into number of moles by dividing by the

molar mass, and then use the coefficients of the chemical equation to find the ratio of moles of

propane to moles of carbon dioxide. After finding the moles of carbon dioxide, multiply by the

molar mass of carbon dioxide to convert moles to grams.

The chemical reaction was found here: http://chemistry.about.com/od/chemicalreactions/ss/10-

Examples-of-Chemical-Reactions-in-Everyday-Life_5.htm

I created the quantities and goal of the problem.

Below is the balanced chemical equation for the synthesis of sulfur dioxide. Given .313moles of

sulfur, calculate how many liters of sulfur dioxide can be created at STP.

S (s) + O2 (g) → SO2 (g)

Using the information given, you multiply the .313 grams of sulfur by the ratio of moles of sulfur

dioxide to moles of sulfur that you can find in the balanced chemical equation. In this case, it is

simply one to one. Now we need to find liters. At STP, 1 mole of a gas is 22.4 liters, so we will

multiply by this number, canceling out moles and getting 7.01 liters of sulfur dioxide gas.

I found the equation here: http://chemistry.about.com/cs/generalchemistry/a/aa072103a.htm

I created the other details of the problem: quantity, what we were trying to find.

Usually, you will have limited quantities of everything. There will be one substance that gets

used up faster than the other substances. This is called the limiting reactant. To find the limiting

reactant, simply use each given quantity to calculate the number of moles of product.

Below is the chemical reaction for the. You have 8.00g of zinc, but you also have 32.0g of

copper sulfate. Find the limiting reactant, and calculate how many grams of zinc sulfate and

copper are produced. After, find how many moles of the other reactant are left.

Zn (s) + CuSO4 (aq) → Cu (s) + ZnSO4 (aq)

Zinc is the limiting reactant, so only 7.78g of copper are produced.

First you must find which substance is the limiting reactant. Take the 8.00g of zinc and use the

molar mass to find the number of moles you have. Taking the number of moles, use the balanced

chemical equation to find the number of moles of copper you have, and convert to grams using

the molar ratio. Now, do the same thing with the 32.0g of copper sulfate, finding moles of copper

sulfate, moles of copper, and then converting to mass. Whichever substance, copper sulfate or

zinc, gave you the least grams of copper is the limiting reactant. Now using this reactant, you can

find the mass of zinc sulfate produced by going through the same steps as with the copper. You

find the number of moles of zinc, the number of moles of zinc sulfate and then convert to grams.

Lastly, you can find the mass of copper sulfate left over from the reaction by the same method.

You would start with 8g of zinc, convert to moles, use the molar ratio to find the number of

moles of copper sulfate and then convert this into grams. Then subtract this mass from the initial

mass to find the number of grams remaining.

I found the reaction here: http://chemistry.about.com/cs/generalchemistry/a/aa072103a.htm

I created the conditions of the problem, the quantities, etc.

When performing experiments in a lab, one way of analyzing error in the lab is by finding the

percent yield of the reaction. Given the amount of substance that reacted, you can calculate the

amount of product that you theoretically should get. Then, you collect experimental data on what

you actually did get.

Below is the reaction for the decomposition of mercury (II) oxide. 5.00g of mercury (II) oxide

was heated in a test tube, and the gas was collected to be 0.201 liters of O2. Find the percent

yield of the reaction.

2HgO (s) + heat → 2Hg (l) + O2 (g)

First you would find the theoretical yield by converting grams of mercury (II) oxide into moles.

Then you would find the number of moles of oxygen and convert to grams. Using this number,

you would then divide .201 by it and multiply by 100. Then you have calculated percent yield.

I found the reaction here: http://chemistry.about.com/cs/generalchemistry/a/aa072103a.htm

I created the conditions of the problem.

Empirical formula: simplest ratio of atoms in a molecule

CH, C2H2 and C6H6 all have the empirical formula of CH. There is one carbon for every one

hydrogen.

You can find empirical formula using percent composition, masses of each atom in the molecule.

Percent Composition

Winds on Venus can reach

up to 450 miles per hour

☼ Assume 100g and find number of grams for each kind of atom. Then use the mass

and convert to moles. Divide each by the smallest number of moles present. If the

resulting moles are decimal numbers, multiply until you get near whole numbers.

If the number is 2.98, simply round up to 3. The number of moles of each atom

are the the subscripts in the empirical formula.

Mass

☼ Instead of assuming 100g, simply use the masses provided. Then follow the same

exact steps as above.

You are given a substance that contains 24.74% potassium, 34.76% manganese, and 40.50%

oxygen. Find the empirical formula.

Assume 100g of substance.

KMnO4

By assuming 100g, you can turn percent into grams easily. Then you take the grams of each and

multiply by the molar mass to find the number of moles of each you have. After you have the

number of moles, divide all of them by the number of moles to et the empirical formula.

I got the compound here:

http://www.syvum.com/cgi/online/mult.cgi/squizzes/chem/compds1.tdf?0

Finding molecular formula is much like finding empirical formula but with a few extra steps.

First find the empirical formula

Find the molar mass of the molecule the empirical formula gives

Find the ratio of the molar mass of the empirical formula and the actual molar mass

Multiply the subscripts of the empirical formula by this ratio

You have a compound that is composed of 82.66% carbon and 17.34% hydrogen. The molar

mass is 58.12g/mol. Find the molecular formula.

Assume 100 g

( )

( ) ( )

2( 2 molc + 5 molH) = 4 molC + 10 molH

C4H10

Assuming 100g, you can again find the moles of each element using the molar mass. Then you

can find the empirical formula using the same steps as before. However, since you are given the

molar mass, you can find the molecular mass. You can find the molar mass of the empirical

formula and divide the given molar mass by this empirical formula molar mass. This will give

you a number which you can then multiply by the empirical formula to get the molecular

formula.

I created the question

Ch. 4

There are several different ways to measure concentration:

Molarity

☼ This unit is M; to find molarity, divide number of moles of the substance by the

number of liters of solution

Molality

☼ This unit is m; to find molality, divide moles of the substance by kilograms of the

solvent used

Mole ratio

☼ There are no units; to find mole ratio, divide the number of moles of substance by

number of moles total

If you have 10.00g of HCl and you dissolve it in water to produce 50mL of solution, find the

molarity.

The formula for molarity is

The air pressure on Venus is 90x

greater than the atmosphere of earth

First, you must convert grams to moles using the molar mass. Then you need to convert

milliliters to liters. Then plug it all into the equation to find molarity.

I created the question.

Given 25mL of .20M H2SO4, you proceed to add 15 mL to the solution to dilute it. Find the

molarity of the new solution.

The formula for molarity is

L

Given the initial molarity and the volume, you must first find the number of moles of H2SO4 in

the solution using the equation for molarity set to moles equals molarity times liters. Then you

must find the new volume of solution by adding together the volumes and converting to liters.

Then simply plug these numbers into the equation for molarity to get the molarity of the diluted

solution.

I created this question.

You have 20.0g of NaCL dissolved into 50.0mL of water. Find the molality. (The density of

water is of course 1.00g/mL)

The equation for molality is

To solve this problem, you must first find the moles of NaCl dissolved. Then you must convert

milliliters of water to grams of water and then to kilograms. Finally, plug in these numbers into

the equation for molality.

I created this question

15.0g of NaHSO4 has been dissolved in 30.0mL of water. Calculate the mole fraction.

First you must determine the number of moles of solute dissolved in the water using the molar

mass. Then find the grams of water in 30.0mL and convert to moles as well. Next plug in these

values into the equation for mole fraction with the denominator the sum of the two numbers,

moles of water and moles of sodium bisulfate, to find mole fraction.

I found the compound here: http://hyperphysics.phy-astr.gsu.edu/hbase/chemical/saltcom.html

I determined the conditions of the problem.

Solubility Rules:

All compounds of Group IA elements (the alkali metals) and the ammonium ion are soluble

All nitrate (NO3-), chlorate (ClO3

-), perchlorate (ClO4

-), and acetate (C2H3O2

-) salts are soluble

Except AgC2H3O2 and KClO4 is only slightly soluble

All chloride (Cl-), bromide (Br

-), and iodide (I

-) salts are soluble

Except for those of Ag+, Pb

2+, and Hg2

2+.

Sulfate compounds are soluble

Except Hg2+

, Pb2+

, Sr2, Ba

2+; and Ca

2+ and Ag

+ are only slightly insoluble

All hydroxide (OH-) compounds are insoluble

Except those of Group I-A (alkali metals) and Ba2+

, Ca2+

, and Sr2+

All compounds are insoluble

Except those of Groups I-A and II-A (alkali metals and alkali earths).

All sulfites (SO3=), carbonates (CO3

=), chromates (CrO4

=), and phosphates (PO4

3-) are insoluble

Except for those of NH4+ and Group I-A (alkali metals)

Strong Acids:

HI

HCl

HBr

HClO3

HClO4

HNO3

H2SO4

Strong Bases:

Venus was named after the

Roman goddess of love.

NaOH

KOH

LiOH

RbOH

CsOH

Ca(OH)2

Ba(OH)2

Sr(OH)2

Net ionic equations

These are when you simplify chemical equations to only those involved in the reaction. Often,

there are “spectator ions” that are unnecessary in the BCE.

To find and weed out spectator ions, follow the solubility rules, and write soluble molecules as

separate ions. Then, you can cancel out ions if they are alone and have the same subscripts and

superscripts on both sides

Write the net ionic equation for the reaction below:

Na+Cl

- + Ag

+NO3

- Ag

+Cl

- +Na

+NO3

-

Na+ + Cl

- + Ag

+ + NO3

+ Ag

+Cl

- + Na

+ + NO3

-

The Na+ and the NO3

+ are just spectator ions in the reaction, and so the net ionic equation is:

Cl- + Ag

+ Ag

+Cl

-

I created this question.

The same would occur if H+Cl

- were substituted for Na

+Cl

-. Since HCl is a strong acid, it

completely ionizes in water, just as NaCl would. It would reduce to the same reaction as above.

H+Cl

- + Ag

+NO3

- Ag

+Cl

- +H

+NO3

-

H+ + Cl

- + Ag

+ + NO3

+ Ag

+Cl

- + H

+ + NO3

-

Cl- + Ag

+ Ag

+Cl

-

I created this question.

Below is the balanced chemical equation of the reaction of barium nitrate and sodium carbonate.

Ba(NO3)2(aq) + Na2CO3(aq) BaCO3(s) + 2 NaNO3(aq)

Given 30.0mL of .0500M barium nitrate, determine the grams of precipitate formed.

( )( )

( ) ( )

First, you convert the milliliters to liters and then plug in liters and molarity to find moles. Using

moles, you can use the BCE to find moles of barium carbonate, which can then be converted to

grams.

I created this problem.

Find the molarity of HCl, if 35mL of .02M NaOH was used to neutralize 20.mL of HCl. Below

is the balanced chemical equation for the reaction between HCl and NaOH:

HCl + NaOH NaCl + H2O

The clouds in Venus’s atmosphere

are sulfuric acid, whose chemical

formula is H2SO4

For this problem, you can use the equation,

, where M is the molarity of the

solution, V is the volume, and F is the chemical equation coefficient.

( )( )( )

( )( )

M2 = .035M

For this equation, you don’t need to convert anything, as long as volume is the same units

throughout. Rearrange the equation to set it equal to M2 and plug in numbers for each variable,

and simplify.

I created this problem.