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

Answer:

C atoms= 13

H atoms= 23

S atoms= 1

O atoms= 4

Explanation:

No. of C moles=

$\frac{56.7 0g}{12 \frac{g}{mol} } = 4.73 \: mol$

No. of H moles=

$\frac{8.419g}{1 \frac{g}{mol} } = 8.419 \: mol$

No.of S moles=

$\frac{11.65g}{32 \frac{g}{mol} } = 0.364 \: mol$

No.of O moles=

$\frac{23.24g}{16 \frac{g}{mol} } = 1.4525 \: mol$

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

$c_{13} \: h_{23} \: s_{1} \: o_{4}$

$c_{13} \: h_{23} \: s_{1} \: o_{4} \times 1 = \\ c_{13} \: h_{23} \: s_{1} \: o_{4}$

To find the final answer

$\frac{Molar \: mass \: of \: real \: formula }{Molar mass of empriracl formula}$

$= \frac{274.5}{274}$

It approximately equals 1

So,

$c_{13} \: h_{23} \: s_{1} \: o_{4} \times 1 = \: \\ c_{13} \: h_{23} \: s_{1} \: o_{4}$