Need helping finding empirical formula

The answer to your question is    C₃H₆O

Explanation:

Process

1.- Express the percents to grams

Carbon = 62 g

Hydrogen = 10.4 g

Oxygen = 27.4 g

2.- Convert the grams to moles

                    12 g of C 1 mol

                    62 g of C  x

                       x = (62 x 1)/12

                       x = 5.17 moles of Carbon

                     1 g of H 1 mol

                    10.4 g       x

                       x = (10.4 x 1) / 1

                       x = 10.4 moles of H

                     16 g of O 1 mol

                      27.4 g of O  x

                      x = (27.4 x 1)/16

                      x = 1.71 moles of O

3.- Divide by the lowest number of moles (1.71)

Carbon   = 5.17/1.71 = 3

Hydrogen = 10.4/1.71 = 6

Oxygen = 1.71/1.71 = 1

4.- Write the empirical formula

                     C₃H₆O

The empirical formula is C3H6O

Explanation:

Step 1: Data given

Suppose mass of comound X = 100.0 grams

Compound X contains:

62.0 % carbon = 62.0 grams

10.4 % hydrogen = 10.4 grams

27.5 % oxygen = 27.5 grams

Molar mass of C = 12.01 g/mol

Molar mass of H = 1.01 g/mol

Molar mass of O = 16.00 g/mol

Step 2: Calculate moles

Mol = mass / molar mass

Mol C = 62.0 grams / 12.01 g/mol

Mol C = 5.16 moles

Mol H = 10.4 grams / 1.01 g/mol

Mol H = 10.3 moles

Mol O = 27.5 grams / 16.00 g/mol

Mol O = 1.72 moles

Step 3: Calculate mol ratio

We divide by the smallest amount of moles

C: 5.16 moles / 1.72 moles = 3

H = 10.3 moles / 1.72 moles =  6

O = 1.72 moles / 1.72 moles = 1

The empirical formula is C3H6O

the cold front replaces the warmer air at the ground level changing the wind to be breezy and denser.there is also an immediate drop in the temperature.

(b) which contains higher levels of vitamin b, oranges or spinach?

explanation:

you can eliminate a, c, and d because they are questions based on opinions.