Hello, Hope,
I answered the same question for another person, so I copied and pasted it here. There are many questions, but all concern the meaning of molar mass and the mole. I'll explain those concepts and then do a couple of the problems for illustration. I'm hoping it will demonstrate how to solve the rest.
The important thing to remember about the mole is that it is just shorthand for a very large number, 6.023x1023, also known as Avogadro's Number. The analogy would be a dozen. When someone says "dozen," you know they mean 12. When someone says "mole," in a science setting, it means 6.023x1023 of something, usually elements and molecules, but I could say "I ate a mole of donuts from Dunkin Donuts" and you'd know I ate 6.023x1023 donuts.
This number has a special importance in science. If one wanted to run a reaction in a manner not to waste molecules, then that person would need to know the actual numbers of atoms going into the reaction. Since hiring someone with tweezers to count them out one at a time seemed like a dead end, early researchers discovered that you could take the mass of whatever you wanted to react and calculate the actual number of molecules by using the "molar mass" of that element or compound.
The molar mass is that element's atomic weight but expressed in grams instead of AMU (atomic mass units). Copper has an AMU of 63.546, so if you measured out 63.546 grams of copper, you'd have 6.023x1023 atoms of copper. The same applies with compounds. Determine the molar mass of a compound, such as H2O by adding the atomic weights of each element every time they appear. Water has 18 for the oxygen and 2x1 for the two hydrogens, for a total of 18, which is expressed as 18 grams/mole. It takes 18 grams of water to have 6.023x1023 molecules of H2O.
Avogadro's number is a common unit, so it is helpful to memorize it.
The other important point is that the definitions of a mole and molar mass include units that can be very helpful in determining the correct calculations. Let's look at the first question you had:
1. Convert 1.29 x 10^24 atoms of HF to grams.
First, calculate the molar mass of HF. From the periodic table we find the atomic weights for H(1.0 amu) and F(19.0 amu). FHF has one of each atom, so add the atomic weights of both to get the atomic mass of HF (1.0 + 19.0 = 20.0 amu). The beauty of the periodic table and the mole is the fact that the atomic mass is also the molar mass when stated with the unit grams/mole. The molar mass of HF is 20.0 grams/mole. This tells us that 20.0 grams of HF contain 6.02 x 1023 molecules of HF. The calculation is the result of two conversions. The first step is to convert 20.0 grams HF to moles HF. We can take the molar mass of (20.0 grams HF)/(mole HF) and invert it, because the inverse is also true. If 20.0 grams HF = 1 moles HF. then 1 mole HF = 20.0 grams HF. Since both the numerator and denominator are equivalent, the conversion factor is equal to "1," and can be inverted, and multiplied by anything, since it is equal to "1." The numerical change is simply due to the conversion of one unit to the other.
The conversions we seek must change the value of atoms HF to grams HF.
(1.29 x 1024 atoms HF)*(1 mole HF/6.02 x 1023 atoms HF)(20.0 grams HF)*(1 mole HF)
Cancel the units first to see if we wind up with the desired unit, grams. atoms and moles HF cancel, leaving just grams HF on top. I get 42.9 grams HF. Is this about what you might predict? Well, we can see that we have about twice as many atoms as we need for Avogadro's number (12.9 x 1023/6.02 x 1023 = around 2). Since each mole of HF has a mass of 20 grams, the value of 42.9 grams seems OK.
Let's try question 3:
3. Convert 2.79g NH3 to atoms
Find the molar mass of NH3 (14 + 3 = 17 amu, or grams/mole). It takes 17 grams of NH4 to produce one mole, or 6.02 x 1023 molecules/(atoms) of NH3. Since we only have 2.79 grams, we will have less than a mole. Use the units to figure how how to find the correct calculation. We start with grams, convert to moles, and then convert the moles to molecules. Watch how the units cancel when the conversion factors are manipulated such that just the unit we want remains.
(2.79 grams NH3)*(1 mole NH3/17.0 grams NH3)*(6.02 x 1023 atoms NH3 /1 mole NH3)
Grams cancel, moles cancel and we are left with molecules(atoms) NH3. I get 9.88 x 1022 molecules of NH3. This is expected - we have less than a mole.
These should help you through the rest of the problems. There is a lot of calculation involved, but the process is the same. WATCH THE UNITS, and you will survive. Just be careful with the calculations. Use a spreadsheet and it will simplify the process.
Bob
