Determine the empirical formula of a compound containing 47.37 grams of carbon, 10.59 grams of hydrogen, and 42.04 grams of oxygen. In an experiment, the molar mass of the compound was determined to be 228.276 g/mol. What is the molecular formula of the compound?

For both questions, show your work or explain how you determined the formulas by giving specific values used in calculations. (10 points)

Mass of C = 47.37g

Mass of H = 10.59g

Mass of O = 42.04g

The total mass of these elements is 100g, taking a proportion of their molar masses.

C = 47.37/12= 3.95

H = 10.59/1 = 10.59

O = 42.04/16= 2.63.

Dividing through with the smallest proportion which is 2.63

C=3.95/2.63 = 1.5

H =10.59/2.63 =4

O = 2.63/2.63= 1

Multiplying through by 2 to get a whole number.

C = 1.5x2 = 3

H= 4x2 = 8

O = 1x2= 2

The empirical formula is C3H6O2

(Empirical formula)n= molecular mass

(C3H8O2)n =228.276

(12x3 +8+16x2)n= 228.276

76n = 228.276

n = 228.276/76

n = 3

Molecular formula = Empirical formula

=(C3H8O2)3 = C9H24O6

The molecular formula is C9H24O6

i had this on my unit activity, i can you if you are a plato student. are you a plato student?

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"