Chemistry Chapter 5 quiz 1

amu

Atomic Mass Units (weight)

A carbon atom weighs 12.01 amu (¹²C)

avogado’s number

6.02E23

6.02E23

1 mole

example:

1 mol C = 6.02E23 atoms = 12.01 g

1 mol H₂O = 6.02E23 atoms = 18.02 g

How many grams are in 3.80 moles of Ca?

3.80 mol Ca (40.08 g Ca / 1 mol Ca) = 152 g Ca

How many moles are 42.50 grams of Al?

42.50 g Al (1 mol Al / 26.98 g Al) = 1.58 moles of Al

How many atoms are in 0.551 grams K?

0.551 g K ( 1 mol K / 39.10 g K) (6.02E23 atoms K / 1 mol K) = 8.48E21 atoms K

Volume changes with …

temperature and pressure (must defien a standard temperature and pressure)

STP

standard temperature and prssure

Standard Temperature

0 C (273 K)

Standard Pressure

1 atmosphere (atm)

At STP, 1 mole of a gas equals …

22.4 L

22.4 L/mol is called molar volume

22.4 L He = 1 mole He = 6.02E23He

How many liters do 0.543 moles of oxygen gas occupy at STP?

0.543 mol O₂(22.4 L O₂ / 1 mol O₂) = 12.2 L O₂

How many moles are 1.25E4 liters of carbon dioxide at STP?

1.25E4 L CO₂ (1 mol CO₂ / 22.4 L CO₂) = 558 mol CO₂

Calculate the number of grams in a 0.088 L sample of fluorine gas at STP?

0.088 L F₂ (1 mol F₂ / 22.4 L F₂) (38.00 g F₂ / 1 mol F₂) = 0.15 g F₂

If given the density of a solid or liquid …

need the density of the substance

density (d)

mass / volume

density units

g/mL

g/cm³

g/L

density of water

1.00 g/mL

Calculate the number of moles of 25.0 mL of hexane C₆H₁₄ (d = 0.6548 g / mL)

25.0 mL C₆H₁₄ (0.6548 g C₆H₁₄ / 1 mL C₆H₁₄) (1 mol C₆H₁₄ / 86.20 g C₆H₁₄) = 0.190 mol C₆H₁₄

solution

a homogeneous mixture

Amounts in solution are expressed as …

molarity (M) = moles of solute / 1 liter of solution

0.75 M NaCl = 0.75 moles of NaCl in 1 L of solution

How many moles of NaCl are in 25.5 mL of 0.75 M NaCl solution

25.5 mL sol (1 L / 1000 mL) (0.75 mol / 1 L sol) = 0.019 mol NaCl

percentage by mass of each element in a compound

find total mass and divide each element by it

If the formula of the compound is given …

percent of mass element = grams of mass element in 1 mol / molar mass (MM) of compound

MM= 3(12.01) +8(1.01) = 44.11 g

% of C = (3(12.01) g /44,11 g) (100%) = 81.68%

Emperical Formula

lowest number ratio

Molecular Formula

normal ratio

Find emperical formula given percentages

25.9% N and 74.1% O

Step 1: assume 100 grams

25.9 g N and 74.1 g O

Step 2: Calculate mols

25.9 g N (1 mol N / 14.01 g N) = 1.85 mol N

74.1 g O (1 mol O / 16.00 g N) = 4.63 mol O

Step 3: divide by smallest mol

N₁O_2.5

Step 4: multiply by same number (If necessary)

N₂O₅

Find molecular formula

27.3% C and 72.7% O (molar mass is 132.03 g / mol)

Step 1: assume 100 grams

27.3 g C and 72.7 g O

Step 2: Calculate mols

27.3 g C (1 mol C / 12.01 g C) = 2.27 mol C

72.7 g O (1 mol O / 16.00 g N) = 4.54 mol O

Step 3: divide by smallest mol

C₁O₂

Step 4: multiply by same number (If necessary)

C₁O₂

Step 5: add molar mass of compound

12.01+2(16.00)=44.01

Step 6: divide given molar mass by the empirical moral mass

132.03/44.01 = 3

Step 7: multiply empirical formula by the qoutient

C₃O₆

reaction (rxn)

one or more substances is changed into one or more new substances

reactants → products

Stoichiometry

the calculations of quantities in rxns

atom

mg

molecule

O₂

formula unit

mgO

BrINClHOF

Always diatomic when by itself

Bromine (Br)

Iodine (I)

Nitrogen (N)

Chlorine (Cl)

Hydrogen (H)

Oxygen (O)

Fluorine (F)

Stoichiometry rules

any starting amount must be converted to moles

then convert moles of one substance into moles of another using mole ratio

then moles of product can be converted into grams, liters, or particles

How many moles of oxygen are formed if 2.39 moles of Al are formed in a rxn

2.39 mol Al (3 mol 0₂ / 4 mol Al) = 1.79

How many moles of Al are formed from 0.65 g of Al₂O₃

0.65 g Al₂O₃ (1 mol Al₂O₃ /101.96 g Al₂O₃) (4 mol Al / 2 mol Al₂O₃ ) = 0.013 mol Al

If 0.033 g Al are formed in this rxn, how many molecules of Al₂O₃ reacted?

0.33 g Al (1 mol Al / 26.98 g Al) (2 mol Al₂O₃ / 4 mol Al) (6.02E23 molecules Al₂O₃ / 1 mol Al₂O₃) = 3.7E20

Limiting Reactant

reactant that gets used up

Excess Reactant

reactant that is left over

When do you need to find the limiting reactant before answering?

when you have amounts of both reactants

When given only one reactant …

you can assume the other is the limiting reactant

What is the unit used in all ice chart values

moles

Steps for Determining Limiting Reactant

1: Calculate the number of moles of each reactant

2: Complete ICE chart

3: Answer the question

You have a sealed vessel that contains 623 mL of oxygen and 25.0 g of sodium. How much sodium oxide can you produce in grams?

Step 1:

623 mL O₂ (1 L / 1000 mL) (1 mol /22.4 L) = 0.0278 mol O

25.0 g Na (1 mol / 22.9 g) = 1.09 mol Na

Step 2:

0.0278 -0.0278 = 0 mol

1.09 - 4(0.0278) = 0.98 mol

0 + 2(0.0278) = 0.0556 mol

Step 3:

0.0556 mol (61.98 g / 1 mol) = 3.45 g 2Na₂O_(s)

Theoretical Yield

the amount you actually get

Actual Yield

The amount of product you are supposed to get

Percent Yield

(Actual yield / Theoretical yield) 100%

Will Percent Yield ever be over 100%

No

If 15.0 g of Nitrogen reacts with 15 g of Hydrogen, 10.5 g of Ammonia is produced. What is the percent yield of this reaction?

(Find moles and fill out ice chart)

1.07 mol NH₃ (Ammonia) (17.04 g NH₃ / 1 mol NH₃) = 18.2 g NH₃ (theoretical)

(10.5 (actual) /18.2) 100% = 57.7%