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?

A) To determine the powder based on the percent composition of only one element, we need to look at the elements present in both powders.

For vanillin (C8H8O3), the elements present are carbon (C), hydrogen (H), and oxygen (O).

For vanilla sugar (C12H22O11), the elements present are carbon (C), hydrogen (H), and oxygen (O) as well.

Since both powders contain the same elements, we cannot differentiate them based on the element composition alone.

B) To determine if the powder should be used for your diabetic friend's coffee, we need to calculate the ratio of carbon dioxide (CO2) produced to the mass of the powder burned.

1. The molar mass of CO2 is 44.01 g/mol.
2. Convert the mass of CO2 collected from kilograms to grams: 1.54x10^-3 kg = 1.54 g.
3. Calculate the moles of CO2 using its molar mass: moles = mass / molar mass = 1.54 g / 44.01 g/mol.
4. Calculate the moles of the powder burned by using the mole ratio between CO2 and the carbon (C) in the powder. From the balanced chemical equation (to be determined in part C), we know that:
- 1 mole of CO2 is produced from the combustion of 1 mole of C.
- The molecular formula for vanillin (C8H8O3) tells us that there are 8 moles of carbon in 1 mole of vanillin.
- The molecular formula for vanilla sugar (C12H22O11) tells us that there are 12 moles of carbon in 1 mole of vanilla sugar.

Now, let's calculate the moles of powder burned for each case:

For vanillin:
moles of powder = moles of CO2 * (moles of C in vanillin / moles of CO2)
= (1.54 g / 44.01 g/mol) * (8 moles C / 1 mole CO2)

For vanilla sugar:
moles of powder = moles of CO2 * (moles of C in vanilla sugar / moles of CO2)
= (1.54 g / 44.01 g/mol) * (12 moles C / 1 mole CO2)

Once you calculate the moles of powder burned for both cases, compare the results to the initial mass of the powder (1.00 g). If the calculated moles of powder are less than 1.00, it means the powder is not pure and contains other substances, so it should not be used for your friend's coffee.

C) The balanced chemical reaction for the combustion of the powder can be determined by examining the elements present and their stoichiometric ratios. Based on the given formulas:

1. For vanillin (C8H8O3):
C8H8O3 + O2 -> CO2 + H2O

2. For vanilla sugar (C12H22O11):
C12H22O11 + O2 -> 12CO2 + 11H2O

Please note that the coefficients in the balanced equations may differ depending on the experimental data.

D) Your friend's statement that "mass wasn't conserved" is not correct. In a closed system, such as the combustion of the powder in a controlled experiment, mass is always conserved. However, it is possible for some of the products (such as water vapor) to escape or be lost during the experiment, leading to a discrepancy in the measured masses. It is important to consider any experimental errors or sources of loss when analyzing the results.