A44.0mg sample of carbon reacts with sulfur to form 122 mg of the compound.
what is the empirical formula of the carbon sulfide? express answer as a chemical formula.

Empirical formula = C₃S₂

Explanation:

Given data:

Mass of carbon = 44.0 mg (44/1000 = 0.044 g)

Mass of sulfur = 122 mg - 44.0 mg = 78 mg = 78/1000 = 0.078 g)

Empirical formula = ?

Solution:

First of all we will calculate the number of moles.

Number of moles of carbon = mass / molar mass

Number of moles of carbon = 0.044 g/ 12.01 g/mol

Number of moles of carbon = 0.0037 mol

Number of moles of sulfur:

Number of moles = mass / molar mass

Number of moles = 0.078 g/ 32,066 g/mol

Number of moles  = 0.0024 mol

Now we will compare the moles:

                C                            :                  S

             0.0037/0.0024        :             0.0024/0.0024

                  1.5                         :               1

C : S = 2(1.5 : 1)

C : S = 3 : 2

Empirical formula = C₃S₂

the answer is c i believe

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml