Question

An organic compound was synthesized and found to contain only c, h, n, o, and cl. it was observed that when a 0.150 g sample of the compound was burned, it produced 0.138 g of co2 and 0.0566 g of h2o. all the nitrogen in a different 0.200 g sample of the compound was converted to nh3, which was had a mass of 0.0238 g. finally, the chlorine in a 0.125 g sample of the compound was converted to agcl. the agcl, when dried, weighed 0.251 g. determine the empirical formula of this compound.

Answer 1

To determine the empirical formula of the compound, we need to calculate the ratios of the different elements present in the compound. The empirical formula of the compound is C7.34H33.06NCl2.08.

Step-by-step explanation:

To determine the empirical formula of the compound, we need to calculate the ratios of the different elements present in the compound. Let's start by calculating the moles of carbon, hydrogen, nitrogen, oxygen, and chlorine using their respective masses. From the given information, we know that the compound is 61.0% carbon, 15.4% hydrogen, 23.7% nitrogen, and 0% oxygen.

Using the molar masses of the elements and their respective masses, we can calculate the moles:

C: 0.150 g * (1 mol/12.01 g) = 0.01249 mol

H: 0.0566 g * (1 mol/1.008 g) = 0.0562 mol

N: 0.0238 g * (1 mol/14.01 g) = 0.0017 mol

O: 0 g * (1 mol/16.00 g) = 0 mol

Next, we need to find the moles of chlorine:

Cl: 0.125 g * (1 mol/35.45 g) = 0.00353 mol

Now, we divide all the moles by the smallest mole value to find the empirical formula:

C: 0.01249/0.0017 = 7.34

H: 0.0562/0.0017 = 33.06

N: 0.0017/0.0017 = 1

O: 0/0.0017 = 0

Cl: 0.00353/0.0017 = 2.08

Therefore, the empirical formula of the compound is

C7.34H33.06NCl2.08

Answer 2

The number of moles is determined by the formula:

number of moles =
`(given mass)/(molar mass)`

Now determining the number of moles:

-
`CO_2`

Molar mass of
`CO_2` is
`44.01 g/mol`

`(0.138 g)/(44.01 g/mol) = 0.00316 mol`

-
`C`

`0.00316 mol CO_2* (1 mole of C)/(1 mole of CO_2) = 0.00316 mol`

-
`H_2O`

Molar mass of
`H_2O` is
`18 g/mol`

`(0.0556 g)/(18 g/mol) = 0.00309 mol`

-
`H`

`0.00309 mol H_2O* (2 mole of H)/(1 mole of H_2O) = 0.00618 mol`

-
`NH_3`

Molar mass of
`NH_3` is
`17.031 g/mol`

`(0.0238 g)/(17.031 g/mol) = 0.00139 mol`

-
`N`

`0.00139 mol NH_3* (1 mole of N)/(1 mole of NH_3) = 0.00139 mol`

`0.200 g` sample contains
`0.00139 mol`. So,

`0.150 g` of sample contains:

`(0.150 g)/(0.200 g)* 0.00139 mol = 0.00104 mole N`

-
`AgCl`

Molar mass of
`AgCl` is
`143.321 g/mol`

`(0.125 g)/(143.321 g/mol) = 0.00175 mol`

-
`Cl`

`0.00175 mol AgCl* (1 mole of Cl)/(1 mole of AgCl) = 0.00175 mol`

`0.125 g` sample contains
`0.00175 mol`. So,

`(0.150 g)/(0.125 g)* 0.00175 mol = 0.0021 mole Cl`

Now writing the formula with their respective moles:

`C_(0.00316)H_(0.00618)N_(0.00104)Cl_(0.00210)`

Dividing with the smallest value of moles:

`C_\frac{{0.00316}}{0.00104}H_{(0.00618)/(0.00104)}N_(0.00104)/(0.00104)Cl_\frac{{0.00210}}{0.00104}`

`C_3H_6NCl_2`

Hence, the empirical formula is
`C_3H_6NCl_2`.