Basic Chemistry (Science)
posted by Chris on .
Determining Chemical Formulas?
This was on my test a few days ago for Basic Chemistry, I received 8/20 and I'd really like to see how to work these problems out. *PLEASE SHOW WORK*. Their are four parts to this, If anyone could answer all four with easy explanations and shown work. I'd greatly appreciate it.
A) A science themed coffee shop offers two white powders for vanilla lovers to add to their coffee. One powder is vanillin (C8H8O3) which offers the flavors of the vanilla bean; The other powder is vanilla sugar containing mostly sucrose (C12H22O11) for sweetness as well as flavor. If you were able to use the percent composition of only one element to identify the powder, which would you choose and why?
B) You best friend is a diabetic and you don't want to add the wrong vanilla flavoring so you take a 1.00-gram sample of one powder and burn it. A sample of carbon dioxide gas has a mass of 1.54x10^-3 kilograms is collected. Your best friend collects water from the combustion with a mass of 0.5791 grams. Should you use this powder for your friend's coffee? Justify your answer.
C) Write the balanced Chemical reaction for the combustion in part B
D) While you and your friend sip the appropriately flavored coffees, your friend gasps in horror at the experiment date from part B "Mass wasn't conserved" Is your friend right? why or why not?
I looked at this yesterday and nothing jumped out at me for A.
B you can do several ways. Perhaps the easiest is to calculate how much CO2 is produced by each. (or you could use H2O if you wish).
mol = grams/molar mass.
convert mols product to mols CO2
Convert mols CO2 to g. g = mols x molar mass.
2C8H8O3 + 17O2 ==>16CO2 x 8H2O
mols vanillin = 1.00g/152 = 0.00658
0.00658 x (16 mols CO2/2 mol C8H8O3) = 0.00658 x (16/2) = 0.0526 g CO2. Can't be it; CO2 was 1.54g. Try the sugar.
C12H22O11 + 12O2 ==> 12CO2 + 11H2O
mol = 1/342 = 0.00292
0.00292 x (12 mol CO2/1 mol C12H22O11) = 0.00292 x 12 = 0.0350
0.0350 mol x 44 = 1.54g. Must be this one.
C. The equations are written in B.
D. As for D I don't know to what data they refer. But you know the answer. Mass WAS conserved so they must have been confusing their numbers.