Fructose is a type of sugar found in fruits and honey. Analysis of a 2.00-g
sample shows that it contains 0.80 g of C, 1.06 g of O, and 0.14 g of H. An independent
analysis shows that fructose has a molar mass of 180 g/mol. What is the molecular
formula for fructose?

The molecular formula is C6H12O6

Explanation:

Step 1: Data given

Mass of the sample fructose = 2.00 grams

Mass of C = 0.80 grams

Molar mass of C = 12.01 g/mol

Mass of O = 1.06 grams

Molar mass of O = 16.0 g/mol

Mass of H = 0.14 grams

Molar mass of H = 1.01 g/mol

Molar mass of the compound = 180 g/mol

Step 2: Calculate moles C

Moles C = 0.80 grams / 12.01 g/mol

Moles C = 0.067 moles

Step 3: Calculate moles O

Moles O = 1.06 grams / 16.0 g/mol

Moles O = 0.066 moles

Step 4: Calculate moles H

Moles H = 0.14 grams / 1.01 g/mol

Moles H = 0.139 moles

Step 5: Calculate the mol ratio

We divide by the smallest amount of moles

C: 0.067 moles : 0.066 moles = 1

O: 0.066 moles / 0.066 moles = 1

H: 0.139 moles / 0.066 moles = 2

For each C atom we have 1 O atom and 2 H atoms

The empirical formula is CH2O

The molar mass of this empirical formula is 30 g/mol

Step 6: Calculate the molecular formula of fructose

We have to multiply the empirical formula of CH2O by n

n = 180 g/mol / 30 g/mol

n = 6

The molecular formula = 6*(CH2O)

The molecular formula is C6H12O6

(newtons first law)

conservation of momentum law states :

momentum before collision = momentum after collision

momentum p (kg-m/s)= mass * velocity

say moving car=1 000 kg, velocity = 10 m/s then p = 1 000*10= 10 000 kg-m/s

say still car = 2 000 kg, velocity = 0 m/s then p = 2 000* 0 = 0 kg-m/s

total momentum prior to collision = 10 000 + 0 = 10 000 kg-m/s

momentum after collision = 10 000 kg-m/s

mass now = 1 000 kg+2 000 kg = 3 000 kg

10 000 = 3 000 kg* velocity m/s

10 000/3 000 = velocity

3.33 m/s = velocity after collision

an impulse is a force applied over time , momentum is exclusive of external forces and a perfect collision is implied