Monday, November 9, 2009

Chem Class- November 9, 2009

Today in class, we went over some questions from last time and continued on dealing with the mole with a few practice problems.

Here's one of the examples we did in class: 250 mL of a gas which is known to contain one sulphur atom and an unknown number of fluorides has a mass of 1.63 g at STP.
- We first had to find the molar mass, which was 0.0112 mol (Try it yourself and see if your answer matches. Remember that the units for molar mass are g/mol)
- Here's a newbie. We then had to come up with the chemical formula. There is one sulphur atom, so that leaves us with this so far: SFx. We need to find out the number of Fluride ions

1 x S
? x F
____
146 g/mol -----> 32.1 g/mol
                           x (19.0 g/mol)
                          ___________
                          146 g/mol   -----> 32.1 + 19x = 146
                                                                  19x = 113.9
                                                                      x = 5.99 or 6 atoms

The final chemical formula is: SF6
_________________________________________________

Here's some questions just for practice:

1) 7.09L of a gas has a mass of 4.34 what is the molar mass at STP?
7.09L x 1mol / 22.4 = 0.3165mol
molar mass = 4.34g / 0.3165mol = 13.71g/mol

2) How many moles are in 25.0 mL of HCN (g) at STP?
# of moles = 0.0250 L x 1 mol/22.4 L = 1.12 x 10 the power of negative 3 moles

3) What is the density of gold if 0.02780 mol of gold has a volume of 0.2836 mL?
Density = 0.02780 mol/0.2836 x 10 to the power of negative 3 litres x 197.0 g/ 1 mol = 1.931 x to the power of 4 g/L

4) How many oxygen atoms are contained in 75.0L of SO3 (g) at STP?
# of O atoms = 75.0L x 1 mol/ 22.4 L x 6.02 x 10 to the power of 23 molecules/ 1 mol x 3 atoms/ 1 molecule = 6.05 x 10 the power of 24 O atoms

5) What is the mass of 2.50 L of PF5(g) at STP?
Mass = 2.50 L x 1 mol/22.4 L x 126. 0 g/1 mol = 1.74 mol


Getting tired of the mole yet? Here's a comic that someone made on the mole (Avogadro's number).

No comments:

Post a Comment