2 Chemistry Questions Regarding Molar Mass and Emprical Formula...Please Help?

I don't need just the answer, I have the answer in the book. I need to know how to get to the answer please! I know how to do each type of problem basically, but nothing to this extent. I've tried a couple of ways to get there but I can't seem to come up with the right answer, If you could explain it as well it would be really appreciated!

1. Vitamin B12, cyanocobalamin, is essential for human nutrition. It contains 4.34 percent cobalt by mass. Calculate the molar mass of cyanocobalamin, assuming that there is one atom of cobalt in every molecule of cyanocobalamin.

2. A complete combustion of a sample of propane produced 2.641 g of Carbon Dioxide and 1.442 g of Hydrogen Dioxide (water) as the only products. Find the empirical formula of Propane.

Answer:
1. From a periodic table, Cobalt weighs 58.93 grams per mole. If the cobalt constitutes 4.34% of the total mass, then you can write the equation 58.93 g = 4.34(x), where x is the total mass of Vitamin B12. Rewriting the equation to solve for x, you get x = 58.93g/0.0434 (remember that 4.34% in decimal form is just 0.043; you shift the decimal point over two places). Therefore x = 1,357.83 grams/mole. A quick check with Wikipedia's page on Vitamin B12 confirms that this is pretty close to the correct actual mass. It's a little bit off, but that's because of the use of significant figures, and the presence of isotopes.

2. The formula for propane is C3H8, but it looks like they're trying to get you to derive it from the combustion information. When you burn hydrocarbons like propane, generally you're taking something that contains carbon and hydrogen, and adding oxygen from the air to form these two products (CO2 and H2O). This is why you need air for things to burn. Oxygen is the "oxidant" in the process.
Okay, so let's start with 2.614 grams of CO2 (carbon dioxide). From a periodic table, C weighs 12.011 g/mole, and O weighs 16.00 g/mole. O2 therefore weighs 16.00 x 2 - 32.00. CO2 as a molecule weighs 12.011 + 32.00 = 44.011 grams/mole. So carbon is 12.011/44.011 = 27.29% of the mass of CO2. That means we had 2.614 x 0.2729 = 0.713 grams of carbon.
Now let's take the water. Hydrogen weighs 1.008 grams/mole, and from the last part we know oxygen weighs 16.00 grams/mole. H2 therefore weighs 1.008 x 2 = 2.016 grams per mole. Water as a molecule weighs 2.016 + 16.00 = 18.016 grams/mole. So in water, hydrogen accounts for 2.016/18.016 = 11.2% of the mass. Since we collected 1.442 grams of water from the reaction, we know that we started with 1.442 x 0.112 = 0.16 grams of hydrogen.
Combining these two pieces of information, we started with 0.71 grams of carbon, and 0.16 grams of oxygen. Now we need to use the formula weights again to get us back to moles from grams. We have 0.71/12.011 = 0.06 moles of carbon, and 0.16/1.008 = 0.16 moles of hydrogen. So our ratio of carbon to hydrogen is 0.06 to 0.16, which is 1 to 2.67. But it doesn't make sense to have a molecular formula of C1H2.67. So you multiply both of the subscripts by integers until you get to whole numbers. If we multiply the whole thing by 2, we get C2H5.33, which is still not a whole number. So let’s try multiplying it by 3. That gives us C3H8. Whole numbers! So this must be the formula for propane.

Good luck!

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