com/in/takeAssignment/takeCovalentActivity dolocatore assignment-take8tak OWLv2|

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com/in/takeAssignment/takeCovalentActivity dolocatore assignment-take8tak OWLv2| Online teaching an...XC ionLocators assignment-tak. Seach.. PH Of 3.40. ook Help e. Safety Tools. 6-JDq.O An unknown compound has the formula C,H,O,. You burn 0.1775 g of the compound and isolate 0.4334 g of CO2 and 0.1774 g of H2O. What is the empirical formula of the compound? If the molar mass is 72.1 g'mol, what is the molecular formula? (Enter the elernents m the order. C,H,N,O) Empirical formula Molecular formula: Try Another Version 9 itern attermpts remaining nswer

Explanation / Answer

1)

Number of moles of CO2 = mass of CO2 / molar mass CO2

= 0.4334/44

= 0.00985

Number of moles of H2O = mass of H2O / molar mass H2O

= 0.1774/18

= 0.0098556

Since 1 mol of CO2 has 1 mol of C

Number of moles of C in CO2= 0.00985

Since 1 mol of H2O has 2 mol of H

Number of moles of H = 2*0.0098556 = 0.0197111

Molar mass of O = 16 g/mol

mass O = total mass - mass of C and H

= 0.1775 - 0.00985*12 - 0.0197111*1

= 0.0395889

number of mol of O = mass of O / molar mass of O

= 0.0395889/16.0

= 0.0024743

Divide by smallest to get simplest whole number ratio:

C: 0.00985/0.0024743 = 4

H: 0.0197111/0.0024743 = 8

O: 0.0024743/0.0024743 = 1

So empirical formula is:CHO

2)

Molar mass of C4H8O = 4*MM(C) + 8*MM(H) + 1*MM(O)

= 4*12.01 + 8*1.008 + 1*16.0

= 72.104 g/mol

Now we have:

Molar mass = 72.1 g/mol

Empirical formula mass = 72.104 g/mol

Multiplying factor = molar mass / empirical formula mass

= 72.1/72.104

= 1

So molecular formula is:CHO

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