A sample of a compound containing only carbon and chlorine contains 1.537 g C and 12.10 g Cl. What is the empi

Answer 1

??? Are you asking for Empirical Formula
If so!!!

C = 1.537/12 = 0.128
Cl= 12.10/35.5 = 0.341
C : Cl :: 0.128 : 0.341

Divide through by smallest ratio

C : Cl :: 0.128/0.128 : 0.341/0.128 :: 1: 2.66
Multiply through by 3 to remove decimals
C : Cl :: 3 : 8
Empirical formula is C3Cl8

NB
If the Empirical formula is the same as the Molecular formula, a possible name is:-
1,1,1,2,2,3,3,3- octochloropropane.

Answer 2

Divide 1.537 by 12 and 12.10 by 35.5, and then find the whole-number ratio between the answers. This is done by dividing the larger one by the smaller one.

Answer 3

1.537 g C = 1.537/12.011 moles of C atoms= 0.128 moles
12.10 g Cl = 12.10/35.457 moles of Cl atoms= 0.341 moles
divide moles of C and of Cl by 0.128
ratio of moles of C to moles of Cl = 1 : 2.66
To get a whole number of moles of Cl, multiply the moles of Cl (and of C) by 3 to get the ratio = 3 : 8
empirical formula C3Cl8

Answer 4

So, find the number of moles of each element.

Mol C = 1.537g C / 12.01 g /mol carbon
=0.1279 mol C

Mol Cl = 12.01 g Cl / 35.45 g/mol Cl
=0.3390 mol Cl

Divide each mole value found by the smallest mole value ( .1279)

So you get .1279 C and 2.65 Cl, then round up to the nearest whole number.

So the formula is CCl3