Question

Sevin, the commercial name for an insecticide used to protect crops such as cotton, vegetables, and fruit, is made from carbamic acid. A chemist analyzing a sample of carbamic acid finds 0.8007 g of carbon, 0.9333 g of nitrogen, 0.2016 g of hydrogen, and 2.133 g of oxygen. Determine the empirical formula for carbamic acid. 7. When 1.916 g of titanium is strongly heated in a stream of pure oxygen, a metal oxide sample weighing 3.196 g results. Calculate the empirical formula of the oxide.

Answer 1

The empirical formula of acetic acid is CH₂O, and the molecular formula is C₂H₄O₂.

Step-by-step explanation:

The empirical formula of acetic acid can be determined by finding the ratio of atoms in the compound. Given that the percentage composition of acetic acid is 39.9% carbon, 6.7% hydrogen, and 53.4% oxygen, we can assume that we have 100g of the compound. Converting the percentages to grams gives us 39.9g of carbon, 6.7g of hydrogen, and 53.4g of oxygen. Next, we need to find the moles of each element by dividing the mass by the molar mass. The molar mass of carbon is 12.01 g/mol, hydrogen is 1.008 g/mol, and oxygen is 16.00 g/mol. Dividing the mass of each element by its molar mass gives us 3.32 mol of carbon, 6.65 mol of hydrogen, and 3.34 mol of oxygen. To find the simplest whole-number ratio of atoms, we divide each mole value by the smallest mole value (3.32 mol in this case). This gives us a ratio of 1:2:1. Therefore, the empirical formula for acetic acid is CH₂O.

The molecular formula of acetic acid can be determined if we know the molar mass of the compound. Given that the molar mass of acetic acid is 60.06 g/mol, we can compare it to the empirical formula mass which is 30.03 g/mol (12.01 g/mol for carbon + 2.016 g/mol for hydrogen + 16.00 g/mol for oxygen). Dividing the molar mass by the empirical formula mass gives us 2. Therefore, the molecular formula of acetic acid is C₂H₄O₂.

Answer 2

Step-by-step explanation:

Ratio of mass of C , N , H and O

= .8007 :0.9333:0.2016:2.133

Ratio of moles of C , N , H and O

= .8007/12 : .9333 / 14 : 0.2016 / 1 : 2.133/16

= .0667 : .0667: .2016 : .1333

= .0667 / .0667 : .0667 / .0667 : .2016 /.0667 : .1333 / .0667

= 1 : 1 : 3: 2

Hence empirical formula = CNH₃O₂

7 .

Weight of titanium Ti = 1.916 g

Weight of oxygen = 3.196 - 1.916 = 1.28 g

Ratio of weight of Ti and O

= 1.916 : 1.28

Ratio of moles of Ti and O

1.916/48 : 1.28/16 [ Molecular weight of Titanium is 48 ]

= .04 : .08

= .04/.04 : .08/.04

= 1 :2 .

Empirical formula

TiO₂