2. the empirical formula of a molecule is ch2o. in an experiment, the molar mass of the molecule was determined to be 360.3 g/mol. use this information to determine the molecular formula of the molecule. show your work.
3. sometimes you are given a measured amount of two or more reactants and asked how much product can be formed. explain what steps you would follow to solve this type of problem.
4. this question has two parts:
a. what is percent yield? explain what it is and how it is calculated.
b. in a laboratory experiment, when 35.0 g mg reacted in excess o2, the percent yield of mgo was 90.0%. what was the actual yield of that experiment?

The answer to your question is:

2.- C₁₂ H₂₄ O₁₂

Explanation:

2.-

Data

CH2O

molar mass = 360.3 g/mol

Molar mass of CH2O = 12 + 2 + 16 = 30g

Divide molar mass given by molar mass obtain

                           x = 360.3/30

                          x = 12

Finally

                      C₁₂ H₂₄ O₁₂

Molar mass = (12 x 12) + (24 x 1) + (16 x 12) = 144 + 24 + 192 = 360 g

3.- First we need to write the complete equation of the reaction and balanced it.

Then, we need to convert the mass given to moles of each compound.

After that, we need used rule of three calculate the amount of products based on the moles of reactants given.

Finally, convert the moles to grams.

4.-

a.- It is a relation between the mass of product obtain in an experiment and the mass of a product obtain theoretically times 100.

b.-

35 g of Mg reacted with excess O2

percent yield = 90%

Actual yield = ?

Formula

Percent yield = (actual yield/theoretical yield) x 100

Equation  

                       2Mg  + O2 ⇒ 2MgO

                      48.62 g of Mg 80.62 g of MgO

                      35g                    x

                     x = 58 g of MgO     (Theoretical yield)

Theoretical yield = 58 g of MgO

Actual yield = percent yield x theoretical yield / 100

                    = 90 x 58 / 100

                   = 52. 23 g

2h2o2--> 2h2o + o2

this reaction is of the spontaneous decomposition of hydrogen peroxide down into water and oxygen.

add 2 molecules of hydrogen peroxide and 2 molecules of water. since oxygen is naturally diatomic, the total number of atoms of each element is now the same on both sides of the equation so it is balanced.

this may not be entirely correct but let me know.