Name: __________________________________________________________
Class Schedule: circle date & time for M/W class only
(1) M/W circle/write your class time 8-11; 11-2; 2-5 _______________________
(2) Fri
Chap 3: Sections 3.5 and 3.7 are the most difficult.
1. Balance the equation below:
__PH3 + __O2  __P2O5 + __H2O is balanced
Answer: 2 PH3 + 4 O2  P2O5 + 3 H2O
2. Calculate the formula weight (in amu) or molar mass (in gram/mole) for calcium nitrate, Ca(NO3)2. Note: The
numerical values are identical but the units are different due to the amount of the substance.
Use 48.0 g of CH4 to answer Q3-6:
3. How many moles of CH4 are there in 48.0 g of CH4? 3 moles CH4
Mole of CH4 = mass/molar mass = 48.0/16.0 = 3.0
4. How many carbon atoms are there in 48.0 g of CH4? 1.806x1024 C atoms
From chemical formula, 1 mole CH4 is composed of 1 mole C atoms and 4 moles of H atoms.
And 1 mole C atoms = 6.02x10 23 C atoms
3 moles of CH4 = 3 mole of C atoms = 3 x (6.02x1023) C atoms = 1.806x1024 C atoms
5. How many hydrogen atoms are there in 48.0 g of CH4? 7.22x1024 H atoms
3 x [4x(6.02x1023)] H atoms = 1.806x1024 C atoms = 7.22x1024 H atoms
6. What is the mass percent (%) for H in CH4? 25%
%H = (total mass of H/Molar mass)x100% = (4/16)x100% = 25%
7. Chemical analysis shows the composition of a compound containing carbon, hydrogen, chlorine, and oxygen, to
be 37.84% C, 2.12% H, 55.84% Cl, and 4.20% O. What is its empirical formula? C12H8Cl6O
The empirical formula is the simplest integral ratio of moles among each atom.
Here, there are four different kinds of atoms, C, H, Cl and O.
Thus mole of C = 37.84/12 = 3.15; mole of H = 2.12/1 = 2.12;
mole of Cl = 55.84/35.45 = 1.58; mole of O = 4.20/16 = 0.26.
Note that as long as one of the moles is not an integer, we have to divide the smallest value among them: here
the smallest value is 0.26. So C : H : Cl : O = 3.15/0.26 : 2.12/0.26 : 1.58/0.26 : 0.26/0.26 = 12.1: 8.1 : 6.1: 1 = 12: 8:
6: 1 (round to whole number), which indicates that the empirical formula contains 12 C, 8 H, 6 Cl and 1 O. Thus, the
empirical formula is written as C12H8Cl6O as 1 is usually not written in the formula.
8. What is the empirical formula for methyl benzoate, a compound used in the manufacture of perfumes, contains
70.57% carbon, 5.94% hydrogen, and 23.49% oxygen? C4H4O
Note: subscripts must be integers. This question is similar to Vitamin C (p.p. 99-100 textbook) that requires
multiplying an integer after dividing the smallest value. That is, for example, you can not simply round 1.5 to 2.
This question is similar to vitamin C question:
Thus mole of C = 70.57/12 = 5.88; mole of H = 5.94/1 = 5.94; mole of O = 23.49/16 = 1.47.
Note that as long as one of the moles is not an integer, we have to divide the smallest value among them: here the
smallest value is 1.47. So C : H : O = 5.88/1.47 : 5.94/1.47 : 1.47/1.47 = 4 : 4.04 : 1. Since 4.04 is very close to 4.00
and thus we can round it to 4.00. So C : H : O = 4 : 4 : 1, which indicates that the empirical formula contains four C,
four H and one O. Thus, the empirical formula is written as C4H4O as 1 is usually not written in the formula.
9. Chemical analysis shows the composition of a compound containing carbon and hydrogen, to be 80.00% carbon
and 20% hydrogen and the molar mass is 30 g. What is its molecular formula? C2H6
10. (Stoichiometry) In the reaction of Al(OH)3 with H2SO4, how many moles of water can be produced If the
reaction is begun with 5.500 mole of Al(OH)3?
2Al(OH)3 + 3H2SO4  Al2(SO4)3 + 6H2O
2____
=
6_____
5.500 mole
=
? mole
There are three methods to solve stoichiometric question:
Method One (Ratio approach): Not in the textbook but video.
5.500 x (6/2) = 16.5 moles of H2O.
Method Two (Road map approach; Figure 3.16 textbook): Apply the road map or say the dimensional analysis.
For Q 11.-13. Limiting Reactant Question:
(Hint: Use stoichiometry twice for two known quantities of reactants; select the answer with the smaller quantity
of the property. If puzzled, think about making sandwiches.)
Consider the reaction of 16.0 g of CH4 with 48.0 g of O2:
CH4 + 2O2  CO2 + 2H2O
11. How many grams of H2O are produced? 27 g
Concept: coeffieient x molar mass = mass
1 (mole) is the coefficient of CH4; 16 (g/mole) is the molar mass of CH4.
2 (mole) is the coefficient of O2; 32 (g/mole) is the molar mass of O2.
2 (mole) is the coefficient of H2O; 18 (g/mole) is the molar mass of H2O.
+
2O2  CO2 +
2H2O
CH4
1x16
=
2x18
16
=
?
If CH4 is the limiting reactant, there will be 36 g H2O produced.
Ratio Method
Using gram directly
CH4 +
Using gram directly
2O2  CO2 +
2H2O
2x32
=
2x18
48
=
??
If O2 is the limiting reactant, there will be 27 g H2O produced.
Because O2 produces the least amount of H2O (can compare by using mole, 1.5 < 2, or grams, 27 < 36), it is the true
limiting reagent and the maximum amount of water produced is 27 grams. The CH4 is the excess reagent.
12. Which is the limiting reactant? O2 Which is the excess reactant? CH4
13. How many grams of the excess reactant left? 4 grams of CH4 left.
To calculate how many grams of the excess reagent were consumed, we apply the ratio method:
2O2  CO2 + 2H2O
CH4 +
1x16
=
2x32
???
=
48
Thus, there are 12 g CH4 consumed and there are (16 – 12) = 4 grams of CH4 left.
This is because # grams excess = # grams original – # grams used.
Or say # moles excess = # moles original – # moles used.
For Q14.-16. Consider that 15.6 grams of benzene (C6H6; molar mass 78 g/mol) is mixed with excess HNO3 to
prepare nitrobenzene (C6H5NO2; molar mass 123 g/mol). After the reaction, there are 13.8 grams of nitrobenzene
produced.
C6H6 + HNO3  C6H5NO2 + H2O
14. What is the theoretical yield?24.6 g
+ HNO3 
C6H6
1x78
15.6
C6H5NO2 + H2O
1x123
?
15. What is the actual yield? 13.8 g
16. What is the percent yield of nitrobenzene, C6H5NO2?
% yield = {13.8 g / 24.6 g} x 100% = 56.10%