1. A compound is composed of only C and H. A sample contains 2.38 g C and 0.200 g H. What is the empirical formula of this compound.
Solution: Divide the mass of C by 12 g/mol and the mass of H by 1 g/mol to get moles.
2.38 0.200 ---- = 0.198 ----- = 0.200 12 1 Note that 0.198 is equivalent to 1.0 ----- --- 0.200 1.0Thus the formula has 1 C to 1 H. 2. A substance has an empirical formula of P2O3. The molar mass is 220 g/mol. What is it's molecular formula?
Solution: determine the mass of the empirical formula.
220 --- = 2 110Multiply the empirical formula by the multiplier:
2(P2O3) = P4O6
3. What is the %C by mass in acetone: C3H6O?
Solution: Determine the total mass.
36 -- x 100 = 62.1 58Reaction: 2Na + Cl2 --> 2NaCl
4. What is the mass of Cl2 needed to react with 4.6g Na?
Solution: Determine moles of Na in 4.6g.
4.6 --- = 0.20 23Note stoichiometry of 2 moles Na to 1 mole Cl2 so:
2 0.20 - = ---- x = 0.1 1 xMultiply the mass of Cl2 by 0.1:
0.1 X 71 = 7.1
5. What mass of NaCl can be produced from 4.6 g Na and an excess of Cl2?
Solution: Use Law of Conservation of Mass
Given: 4.6 g Na From 4: 7.1 g Cl2 --- 11.7 g NaCl6. A reaction mixture contains 5.0 mol Na and 4.0 mol Cl2. How many moles of NaCl can be produced?
Solution: look at reaction stoichiometry.
Setup up ratio: 2 5 - = - x = 5 2 x7. How many moles of which reagent will be left unreacted from the mixture in the previous question?
Solution: Look at reaction stoichiometry.
Setup up ratio: 2 5 - = - x = 2.5 1 x2.5 moles used so 4.0 - 2.5 = 1.5 mol Cl2 left over.
8. What mass of NH3 will be produced from the reaction
Solution: look at stoichiometry.
14g 12 --- = 0.5 mole -- = 6 moles 28g 2Note have 6:0.5 which is 12 mol H2 to 1 mol N2 but reaction only needs 3:1. There is excess H2.
1 N2 0.5 N2 ----- = ------ x = 1 NH3 2 NH3 x NH3Determine mass of 1 NH3: 14 + 3(1) = 17 g
9. For the reaction:
Solution: divide actual yield by theoretical (calculated) yield and multiply by 100.
162g ---- x 100 = 88% 185g10. What is the molarity of glucose solution if 54 g glucose, C6H12O6 (180g/mol) is dissolved in enough water to make 0.500 L of solution?
Solution: molarity is:
moles ----- LFirst convert 54g to moles by dividing by the molar mass.
54g -------- = 0.30 180g/molDivide moles by L to get molarity:
0.30 moles ----------- = 0.60 M 0.500 L11. A solution is prepared by diluting 25 mL of 6.0 mol/L HCl to 250 mL with water. What is the concentration of the dilute solution?
Solution: Use m1v1 = m2v2
Reaction stoichiometry is 1 mole to 1 mole.
Solution: determine stoichiometry.
100mL --------- x 0.50 mol/L NaOH = 0.05 mol OH- 1000 mL/L 200 mL --------- x 0.20 mol/L MgCl2 = 0.02 mol Mg2+ 1000 mL/L Ratio is: 1 Mg2+ ------ 2 OH- We have: 0.02 Mg2+ 1 --------- which is --- 0.05 OH- 2.5There is more OH- than needed to use up all the Mg2+ so Mg2+is limiting. Since Mg2+ comes from MgCl2, MgCl2 is the limiting reagent.
14. Which of the following is considered a weak acid?
Solutions: organic (based on C) are always weak acids. Since only one answer can be correct, it must be HCH3COO.
15. Which of the following solutions will form a precipitate when mixed?
Solution: Use solubility rules.
A) all chlorides (except noted) and nitrates are soluble. No precipitate.
B) All carbonates (except noted) are insoluble. Insoluble means doesn't dissolve. Doesn't dissolve means can't be in solution. Can't be in solution means it precipitates.
C) All sulfates (except noted) and all chlorides are soluble.
D) This is a noted carbonate and not a noted sulfate.
E) All chlorides (except noted) are soluble. All hydroxides except noted are insoluble. This is a noted hydroxide and is soluble.
16. Which of the following sets of coefficients properly balance the following precipitation reaction?
In theory: B. the bulb glows dimply because there are relatively low concentrations of ions in solution.
In practice: D. an current flows because a key property of a weak acid is that it does not produce enough ions to carry the electricity through the solution.
18. The following reaction occurs:
This is an acid (H+) base (OH-) reaction. The participating ions are H+ and OH- so:
Solution: Water is polar which gives it attractive forces that pull ions into solution.
20. Which of the following forms of matter has a volume that is highly dependent upon temperature and pressure?
A) solid - doesn't change size much when heated, cooled, pressed on. NOT!
B) liquid - doesn't change size much when heated, cooled,pressed on. NOT!
C) gas - Boy! does this one get compressed!!!! RIGHT ON!
D) colloid - just a really small particle.
21. Which of the following is a unit of pressure?
Solution: learn units of pressure
22. Which instrument is used to measure gas pressure?
Solution: watch the weather channel!!!!!
Solution: Write the Ideal Gas Law.
R is a constant. V is on same side so can't be directly proportional.
Only options are n and T. N isn't a choice.
24. According to the Ideal Gas Law, volume is inversely proportional to ________.
Solution: What can't be!
R is a constant. N and T are on opposite side of = so V is directly proportional to them.
Only option is P
25. An automobile tire at a pressure of 2.38 atm at 298K is heated to 360K. If the volume of the tire is constant, what is the new pressure?
Solution: Initial/Final Ideal Gas Law
P1V1 n1RT1 ---- = ----- P2V2 n2RT2Volume - constant, R - constant, n is constant since air is not moving in or out of tire.
P1 T1 -- = -- P2 T2Plug in values:
2.38 298 ---- = ---- x 360
Solution: sue Ideal Gas Law
Set up ratio.
2 x - = --- x = 8.8L 1 4.428. What volume will 128.2 g SO2(g) occupy at a pressure of 500 torr and a temperature of 127C?
This one takes knowledge and effort.
Need molar mass of SO2. 32 + 2(16) = 64 g/mol
128.2 g ------- SO2(g) = 2 moles so n = 2.0 64 g/molT must be in K and K = C + 273, so K = 127 + 273 = 400 K
Pressure must be in atm so from given info on first page:
1 atm 1 mm Hg 500 torr x --------- x -------- = 0.657 atm 760 mm Hg 1 torrPlug into Ideal Gas Law:
Solution: determine molar mass of HBr = 1 + 79.9 = 80.9
Convert C to K, K = 14 + 273 = 287 K
PM From front page Use D = ---- RT Plug in values: (1.5 atm)(80.9 g/mole) D = --------------------------- (0.0821 L atm/mole K)(287K) D = 5.15 g/L = 5.2 g/l30. Which if the following gases has the highest density at STP?
Solution: Since P, R, and T are constant only molar mass determines density. THe one with the greatest molar mass is the answer.
Need mole fraction so first determine total number of moles.
1 XA = - 4Last using partial pressure equation from front page of test.