GexC Assignment/takeCovalentActivity.do?locator-assignment-take&takeAssignmentSe

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GexC Assignment/takeCovalentActivity.do?locator-assignment-take&takeAssignmentSessionlocator-assignm er Use the References to access important values if needed for this q A compound is found to contain 39.99 % carbon , 6.727 % hydrogen , and 53.28 % oxygen by mass. To answer the question, enter the elements in the order presented above. QUESTION 1: The empirical formula for this compound is QUESTION 2: The molar mass for this compound is 90.09 gmol. The molecular formula for this compound is Submit Answer Retry Entire Group 8 more group attempts remaining

Explanation / Answer

1)

we have mass of each elements as:

C: 39.99 g

H: 6.727 g

O: 53.28 g

Divide by molar mass to get number of moles of each:

C: 39.99/12.01 = 3.3297

H: 6.727/1.008 = 6.6736

O: 53.28/16.0 = 3.33

Divide by smallest to get simplest whole number ratio:

C: 3.3297/3.3297 = 1

H: 6.6736/3.3297 = 2

O: 3.33/3.3297 = 1

  

So empirical formula is:CH2O

Answer: CH2O

2)

Molar mass of CH2O,

MM = 1*MM(C) + 2*MM(H) + 1*MM(O)

= 1*12.01 + 2*1.008 + 1*16.0

= 30.026 g/mol

Now we have:

Molar mass = 90.09 g/mol

Empirical formula mass = 30.026 g/mol

Multiplying factor = molar mass / empirical formula mass

= 90.09/30.026

= 3

So molecular formula is:C3H6O3

Answer: C3H6O3

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