Question

Complete each of the following calculations, showing all work. You must complete the calculations in the order they appear to solve all of the problems. This Web Lab has to be completed and turned in by November 25^th. PLEASE SHOW ALL WORK.

A bottle of vintage port wine has a volume of 750 ml and contains 18% ethanol (C₂H₆O) by volume. Ethanol has a density of 0.789 g/ml. At 20 ^OC port wine has a density of 0.990 g/ml. The alcohol in port wine is made when grape sugar (C₆H₁₂O₆) undergoes fermentation (no oxygen) to ethanol and carbon dioxide. The weight of 150 grapes is 1.5 lb. and contains 26 g of grape sugar (1 ton = 2,000 lb.)

a)Calculate the percent concentration of ethanol by mass. (You have the volume of the wine. 18% of that is ethanol. Find the volume of ethanol. Convert the original volume of wine to mass using the density of the wine. Convert the volume of ethanol to mass using the density of ethanol. Mass percent is mass of the part divided by the mass of the whole. The ethanol is the part, and the wine is the whole.)

b)Write the balanced chemical equation for the fermentation of grape sugar. ( Grape sugar decomposes into ethanol and carbon dioxide.)

c) How many grams of grape sugar are required to produce one bottle of port wine? (Use stoichiometry. From part a you know the grams of ethanol in one bottle of wine. Convert from grams of ethanol to moles of ethanol, then use the balanced chemical reaction from part b to convert from moles of ethanol to moles of grape sugar. Then convert from moles of grape sugar to grams of grape sugar.)

d) How many grapes are needed to make one bottle of port wine? (If you have 26 grams of grape sugar for every 150 grapes then you can convert from grams to grapes.)

e) How many bottles of port wine can be produced from 1.0 ton of grapes? (Convert from tons of grapes to pounds of grapes, 1.5 pounds of grapes give you 26 g of grape sugar, part c gives you how many grams of grape sugar you need for one bottle of wine.)

Answer 1

Ethanol in port wine

a) Volume of ethanol in port wine = 18 x 750 ml/100 = 135 ml

mass of ethanol in wine = 135 ml x 0.789 g/ml = 106.5 g

mass of port wine = 750 ml x 0.990 g/ml = 742.5 g

So,

mass% ethanol in port wine = 106.5 g x 100/742.5 g = 14.34%

b) Balanced chemical equation for fermentation of grape sugar

C6H12O6 ---> 2C2H5OH + 2CO2

c) moles of ethanol in wine = 106.51 g/46.07 g/mol = 2.312 mol

moles of grape sugar needed = 2.312 mol/2 = 1.156 mol

grams of grape sugar needed = 1.156 mol x 180.156 g/mol = 208.26 g

d) We have 26 g grape sugar in 150 grapes

So for 208.26 g grape sugar we would need = 208.26 g x 150 grapes/26 g = 1201.5 grapes

e) 1.0 tons of grapes = 907185 g grapes

1201.5 grapes needed to prepare 1 bottle of wine,

So,

number of bottle of wines prepared = 907185/1201.5 = 755 bottles