Question

Would you please help need asap

The analysis of a compound gives the following percent composition by mass:
C: 56.70 percent; H: 8.419 percent; S: 11.65 percent; O: 23.24 percent. What is its molecular formula, given that its molar mass is 275.4 g?

C?H?S?O?
C atoms
H atoms
S atoms
O atoms

Answer 1

C atoms= 13

H atoms= 23

S atoms= 1

O atoms= 4

Step-by-step explanation:

No. of C moles=

`(56.7 0g)/(12 (g)/(mol) ) = 4.73 \: mol`

No. of H moles=

`(8.419g)/(1 (g)/(mol) ) = 8.419 \: mol`

No.of S moles=

`(11.65g)/(32 (g)/(mol) ) = 0.364 \: mol`

No.of O moles=

`(23.24g)/(16 (g)/(mol) ) = 1.4525 \: mol`

So the ratios are,

C : H : S : O

4.73 : 8.419 : 0.364 : 1.4525

13 : 23 : 1 : 4

So the empirical formula will be

`c_(13) \: h_(23) \: s_(1) \: o_(4)`

`c_(13) \: h_(23) \: s_(1) \: o_(4) * 1 = \\ c_(13) \: h_(23) \: s_(1) \: o_(4)`

To find the final answer

`(Molar \: mass \: of \: real \: formula )/(Molar mass of empriracl formula)`

`= (274.5)/(274)`

It approximately equals 1

So,

`c_(13) \: h_(23) \: s_(1) \: o_(4) * 1 = \: \\ c_(13) \: h_(23) \: s_(1) \: o_(4)`