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algol [13]
1 year ago
14

certain alcohol contains only three elements, carbon, hydrogen, and oxygen. Combustion of a 40.00 gram sample of the alcohol pro

duced 76.40 grams of CO2 and 46.96 grams of H2O. What is the empirical formula of the alcohol? Express your answer as a chemical formula.
Chemistry
1 answer:
seraphim [82]1 year ago
3 0

Answer:

The answer to your question is:    C₂H₆O₁   = C₂H₆O

Explanation:

Data

CxHyOz

mass = 40 g  produced 76.40 g of CO2

                                       46.96 g of H2O

Empirical formula = ?

Process

                         CxHyOz    + O2    ⇒    CO2  + H2O

                         44g of CO2 --------------------  12 g of Carbon

                         76.40 g of CO2 --------------- x

                          x = 20.84 g of Carbon

                         12 g of Carbon ---------------  1 mol

                         20.84 g of C    ---------------   x

                         x = (20.84 x 1) / 12

                         x = 1.74 mol of Carbon

                        18 g of H2O --------------------  2 g of H

                        46.96 g of H2O --------------   x

                        x = (46.96 x 2) / 18

                        x = 5.22 g of H

                        1 g of H ------------------------  1 mol of H

                        5.22 g of H -------------------   x

                        x = 5.22 mol of H

Mass of Oxygen = 40 - 20.84 - 5.22g

                           = 13.94 g

                         16 g of Oxygen ----------------  1 mol

                         13.94 g of O --------------------   x

                         x = 0.87 mol of O

Divide by the lowest number of moles

Carbon = 1.74 / 0.87 = 2

Hydrogen = 5.22 / 0.87 = 6

Oxygen = 0.87 / 0.87 = 1

                           

Empirical formula

                                 C₂H₆O₁   = C₂H₆O

                         

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7 0
1 year ago
A 32.5 g piece of aluminum (which has a specific heat capacity of 0.921 J/g°C) is heated to 82.4°C and dropped into a calorimete
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m(aluminium)*c(aluminium)*ΔT(aluminium) = -m(water)*c(water)*ΔT(water)

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⇒with ΔT(aluminium) = the change of temperature of aluminium = 24.2 °C - 82.4 °C =  -58.2 °C

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A 1.0 g sample of a cashew was burned in a calorimeter containing 1000. g of water, and the temperature of the water changed fro
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The correct answer is option C.

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3.0 g sample of a marshmallows  :

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Heat released on  combustion of 1.0 gram of marshmallows -13,946.67 J.

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Let's convert the moles to mass

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4 0
1 year ago
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