11) A 1.357 gram sample of a compound containing only carbon, hydrogen, and oxygen was burned in excess oxygen gas. The combustion produced 1.989 grams of carbon dioxide and 0.8143g of water .
A. Find the empirical formula of the compound
B. if the molar mass of the is found to be 60.06 g/mol what is the molecular formula

The answer to your question is below

Explanation:

Data

mass of the sample = 1.357 g

mass of carbon dioxide = 1.989 g

mass of water = 0..8143 g

Process

-Moles of Carbon

Molar mass of CO₂ = 44 g

                         44 g of CO₂ 12 g of C

                          1.989 g of CO₂  x

                          x = (1.989 x 12) / 44

                          x = 0.542 g of C

                         12 g of C 1 mol

                         0.542 g of C  x

                         x = (0.542 x 1) / 12

                        x = 0.0452 moles

-Moles of Hydrogen

Molar mass of water = 18 g

                         18 g of H₂O 2 g of H

                         0.8143 g of H₂O x

                         x  = (0.8143 x 2) / 18

                        x = 0.0904 g of H

                          1 g of H 1 mol of H

                         0.0904 g of H  x

                         x = (0.0904 x 1) / 1

                        x = 0.0904 moles of H

-Moles of Oxygen

Mass of Oxygen = 1.357 - 0.542 - 0.0904

                           = 0.7246 g

                          16 g of O 1 mol

                          0.7246 g of O  x

                          x = (0.7246 x 1) / 16

                          x = 0.04528

2.- Divide by the lowest number of moles

Carbon = 0.0452/0.0452 = 1

Hydrogen = 0.0904/0.0452 = 2

Oxygen = 0.0452 /0.0452 = 1

3.- Write the empirical formula

                                 CH₂O

4.- Find the molecular formula

-Molar mass CH₂O = 12 + 2 + 16

                               = 30 g

-Divide the molecular mass by the molar mass   60/30 = 2

-Write the molecular formula

                           2(CH₂O) =  C₂H₄O₂                                        

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explanation:

1.consider the following balanced equation and answer parts a through. show calculations for each part.  

3 ag(s) + 4 hno3(aq) → 3 agno3(aq) + no(g) + 2 h2o(l)  

a. calculate the number of moles of agno3 produced by the reaction of 3 moles of ag with excess hno3. if the number to the left of the compound represents the number of moles of the compound which are present, then 3 moles of ag react with 4 moles hno3 to produce 3 moles agno3 (aq).

so the answer to a would be 3 moles.

b. calculate the number of moles of agno3 produced by the reaction of 108 grams of ag with excess hno3.  

i'm not sure how to plug in the grams of silver into an equation using moles to find how many moles are produced.

c. calculate the number of grams of agno3 produced by the reaction of 108 grams of ag with excess hno3  

d. calculate the number of moles of no produced as a byproduct of the reaction of 108 grams of ag with excess hno3.  

e. calculate the number of grams of no produced as a byproduct starting with 108 grams of ag and excess hno3.  

explanation: