si 11/13

redox

100

MnO2 --------> Mn2O3 

reduction or oxidation?

reduction

200

NaH

oxidation number 

-1

300

NH3 --------> NO2

reduction or oxidation

oxidation

400

Pb SO4

oxidation number 

2

500

O2 --------> O2

oxidation or reduction

reduction

empirical and molecular

100

Find the empirical formula of a compound that is 48.38% carbon, 8.12% hydrogen, and 53.5% oxygen by mass.

1. 
   
2. Get the mass of each element by assuming a certain overall mass for the sample (100 g is a good mass to assume when working with percentages).
   
   Remember that percentages are a ratio multiplied by 100. You must convert percentages back to their decimal value before working with them.
   

   (.4838) (100 g) = 48.38 g C

   (.0812 ) (100 g) = 8.12 g H

   (.5350) (100 g) = 53.38 g O

3. Convert the mass of each element to moles of each element using the atomic masses.
   
   (48.38 g C) (1 mol/ 12.10 g C) = 4.028 mol C

   (8.12 g H) (1 mol/ 1.008 g H) = 8.056 mol H

   (53.38 g O) (1 mol/ 16.00 g O) = 3.336 mol O

4. Find the ratio or the moles of each element by dividing the number of moles of each by the smallest number of moles.
   
   There are fewer moles of oxygen than any other element, so we will assume one mole of oxygen to establish the ratios.
   
   (3.336 mol O/ 3.336) = 1 mol O

   (4.028 mol C/ 3.336) = 1.2 mol C

   (8.056 mol H/ 3.336) = 2.4 mol H

5. Use the mole ratio to write the empirical formula.
   
   The mole ratio did not turn out to be whole numbers. Since the we cannot have partial atoms in the empirical formula, a multiplication factor must be applied to get whole numbers. In this case, 5 is the factor we need.

   (1 mol O) (5) = 5 mol O

   (1.2 mol C) (5) = 6 mol C

   (2.4 mol H) (5) = 12 mol H
   
   Now, the ratios are whole numbers, and we can write the empirical formula:

C6H12O5

200

Find the empirical fomula for the oxide that contains 42.05 g of nitrogen and 95.95 g of oxygen.

NO2

300

A compound has an empirical formula of ClCH2 and a molecular weight of 98.96 g/mol. What is its molecular formula?

Cl2C2H4

400

A compound is 75.46% carbon, 4.43% hydrogen, and 20.10% oxygen by mass. It has a molecular weight of 318.31 g/mol. What is the molecular formula for this compound?

empirical:

C10H7O2 

molecular: C20H14O4 

500

Calculate the percent by mass of phosphorus and chlorine in phosphorus trichloride (PCl3).

1. Assume one mole of phosporus trichloride. Find the moles of phosphorus and moles of chlorine.
   
   (1 mol PCl3) (1 mol P/ 1 mol PCl3) = 1 mol P

   (1 mol PCl3) (3 mol Cl/ 1 mol PCl3) = 3 mol Cl
   
   

2. Convert mole of each element to mass of each element.
   
   (1 mol P) (30.97 g P/ 1 mol P) = 30.97 g P

   (3 mol Cl) (35.45 g Cl/ 1 mol Cl) = 106.4 g Cl
   
   

3. Calculate the percent by mass of each element.
   
   Mass of sample = 30.97 g + 106.4 g = 137.4 g

   Percent P = [(30.97 g) / (137.4 g)] (100%) = 22.54%

   Percent Cl = [(106.4 g) / (137.4 g)] (100%) = 77.44%

net ionic

100

AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq) 

provide the complete ionic and net ionic eqaution

Ag+(aq) + NO3–(aq) + Na+(aq) + Cl–(aq) → AgCl(s) + Na+(aq) + NO3–(aq) 

Ag+(aq) + Cl–(aq) → AgCl(s)

200

CuCl2(aq) + K3PO4(aq) → KCl(aq) + Cu3(PO4)2(s) 

balance eqaution, give ionic and net ionic

balanced: 3CuCl2(aq) + 2K3PO4(aq) → 6KCl(aq) + Cu3(PO4)2(s) 

complete ionic:

3Cu2+(aq) + 6Cl−(aq) + 6K+(aq) + 2PO43−(aq) → 6K+(aq) + 6Cl−(aq) + Cu3(PO4)2(s)

Net Ionic Equation: 3Cu2+(aq)+2PO43−(aq)→Cu3(PO4)2(s) 

300

. All salts of nitrates, chlorates and acetates are soluble 

true or false

true

400

MnCl2 (aq) + (NH4)2CO3 (aq) --> MnCO3 (s) + 2 NH4Cl (aq)

Total Ionic: Mn2+ (aq) + 2 Cl¯ (aq) + 2 NH4 + (aq) + CO3 2- (aq) ---> MnCO3 (s) + 2 NH4 + (aq) + 2 Cl¯ (aq) 

Net Ionic: Mn2+ (aq) + CO3 2- (aq) --> MnCO3 (s)

500

3 Ba(NO3)2 (aq) + 2 (NH4)3PO4 (aq) ---> Ba3(PO4)2 (s) + 6 NH4NO3 (aq

Total Ionic: 3 Ba2+ (aq) + 6 NO3¯ (aq) + 6 NH4 + (aq) + PO4 3- (aq) --> Ba3(PO4)2 (s) + 6 NH4 + (aq) + 6 NO3¯ (aq) 

Net Ionic: 3 Ba2+(aq) + 2 PO4 3- (aq) --> Ba3(PO4)2 (s)

nomenclature

100

vanadium (IV) carbonate

V(CO3)2

200

Fe(NO3)3

Iron(III) nitrate

300

Al(CN)3

Aluminium cyanide

400

diboron tetrabromide

B2Br4

500

magnesium sulfate heptahydrate

MgSO4 · 7H2O

stoichiometry

100

How many moles of I2 are required to react with 0.429 mol of Al according to the following equation?

2Al+3I2⟶2AlI3

0.644 mol I2

200

How many carbon dioxide molecules are produced when 0.75 mol of propane is combusted according to this equation?

C3H8+5O2⟶3CO2+4H2O

1.4×10^24 CO2 molecules

300

What mass of sodium hydroxide, NaOH, would be required to produce 16 g of the antacid milk of magnesia [magnesium hydroxide, Mg(OH)2] by the following reaction?

MgCl2(aq)+2NaOH(aq)⟶Mg(OH)2(s)+2NaCl(aq)

22g NaOH

400

What mass of oxygen gas, O2, from the air is consumed in the combustion of 702 g of octane, C8H18, one of the principal components of gasoline?

2C8H18+25O2⟶16CO2+18H2O

2.46×10^3gO2

500

what does M stand for?

mol/ L