Can someone help me figure out the empirical formula of a tin-oxygen compound?

Question

Objective:

To determine the empirical formula of a tin-oxygen compound.

Procedures:

In a fume hood (for safety reasons), measure about 3g of tin in an evaporating dish and add 5 ml of nitric acid to the tin in the dish. Allow the gas to escape and then evaporate the water by heating the dish over a Bunsen burner.

The remaining compound is a tin-oxygen compound.

Sn + HNO3 --> NO2 + H2O + tin/oxygen compound

Data:

Mass of tin in dish = 3.00 g

Mass of tin-oxygen product after heating = 3.80 g

Calculations:

1. Calculate moles of tin =

2. Calculate moles of oxygen in tin-oxygen compound =

3. Molar ratio of tin to oxygen =

4. Empirical formula =

5. Name the compound _______.

6. What is the difference between an empirical and a molecular formula?

7. What are some sources of experimental error?

Thanks SO much in advance!!

Answer 1

Nope.  Can't help you.  Kthxbai.

Answer 2

Mass of tin: 3.0 g. Moles of tin: 3.00 divided by molar mass Sn (why?)

Mass of O: 0.8 g. Moles of O: 0.80 divided by ...  (how did I know?)

Then you can calculate the ratio.

Divide by the smaller and you should get a whole number or very close to it, or a simple fraction. If you get a fraction, multiply through to get rid of it.

Take it from there. The rest is just book-work.

Answer 3

find moles using molar masses:

answer to #1 is 3.00 g Sn @ 118.7 g/mole = 0.0253 moles Sn

answer to # 2. is 0.80 g O @ 16 g/ mol = 0.050 moles of O

3. Molar ratio of tin to oxygen = 1:1

4. Empirical formula = s****.>
5. Name the compound Tin (II) oxide

6. What is the difference between an empirical and a molecular

formula? an empirical formula is the simplest whole # ratio of atoms in the compound, where as Molecular formula has the actual # of atoms in the molecule:

water: H2O is both its empirical & molecular formula

hydrogen peroxide: empirical is HO, while molecular is H2O2

7. What are some sources of experimental error?

loss of sample due to splattering in the hood during evaporation, (leads to a lower calculated grams of total compound & also  lower grams of oxygen than we should have

sometimes the final compound is not totally dried, which results in a higher calculated weight for the compound & for the oxygen