The solute has to be hydrophilic, (water loving).
Given the percentage composition of HC as C → 81.82 % and H → 18.18 %
So the ratio of number if atoms of C and H in its molecule can will be:
C : H = 81.82 12 : 18.18 1 C : H = 6.82 : 18.18 = 6.82 6.82 : 18.18 6.82 = 1 : 2.66 ≈ 3 : 8
So the Empirical Formula of hydrocarbon is:
C 3 H 8
As the mass of one litre of hydrocarbon is same as that of C O 2 The molar mass of the HC will be same as that of C O 2 i.e 44 g mol
Now let Molecular formula of the HC be ( C 3 H 8 ) n
Using molar mass of C and H the molar mass of the HC from its molecular formula is:
( 3 × 12 + 8 × 1 ) n = 44 n So 44 n = 44 ⇒ n = 1
Hence the molecular formula of HC is C 3 H 8
Does that help?
Answer:
0.00050553
Explanation:
when the power of ten is negative, move the decimal to the left
hope this helped!
Answer:
![\boxed {\boxed {\sf About \ 2.127 \ moles \ of \ FeCl_3}}](https://tex.z-dn.net/?f=%5Cboxed%20%7B%5Cboxed%20%7B%5Csf%20About%20%5C%202.127%20%5C%20moles%20%5C%20of%20%5C%20FeCl_3%7D%7D)
Explanation:
To convert from moles to grams, the molar mass must be used.
1. Find Molar Mass
The compound is iron (III) chloride: FeCl₃
First, find the molar masses of the individual elements in the compound: iron (Fe) and chlorine (Cl).
There are 3 atoms of chlorine, denoted by the subscript after Cl. Multiply the molar mass of chlorine by 3 and add iron's molar mass.
- FeCl₃: 3(35.45 g/mol)+(55.84 g/mol)=162.19 g/mol
This number tells us the grams of FeCl₃ in 1 mole.
2. Calculate Moles
Use the number as a ratio.
![\frac{162.19 \ g \ FeCl_3}{1 \ mol \ FeCl_3}](https://tex.z-dn.net/?f=%5Cfrac%7B162.19%20%5C%20g%20%5C%20FeCl_3%7D%7B1%20%5C%20mol%20%5C%20FeCl_3%7D)
Multiply by the given number of grams, 345.0.
![345.0 \ g \ FeCl_3 *\frac{162.19 \ g \ FeCl_3}{1 \ mol \ FeCl_3}](https://tex.z-dn.net/?f=345.0%20%5C%20g%20%5C%20FeCl_3%20%2A%5Cfrac%7B162.19%20%5C%20g%20%5C%20FeCl_3%7D%7B1%20%5C%20mol%20%5C%20FeCl_3%7D)
Flip the fraction so the grams of FeCl₃ will cancel.
![345.0 \ g \ FeCl_3 *\frac{1 \ mol \ FeCl_3}{162.19 \ g \ FeCl_3}](https://tex.z-dn.net/?f=345.0%20%5C%20g%20%5C%20FeCl_3%20%2A%5Cfrac%7B1%20%5C%20mol%20%5C%20FeCl_3%7D%7B162.19%20%5C%20g%20%5C%20FeCl_3%7D)
![345.0 *\frac{1 \ mol \ FeCl_3}{162.19 }](https://tex.z-dn.net/?f=345.0%20%2A%5Cfrac%7B1%20%5C%20mol%20%5C%20FeCl_3%7D%7B162.19%20%7D)
![\frac{345.0 \ mol \ FeCl_3}{162.19 }](https://tex.z-dn.net/?f=%5Cfrac%7B345.0%20%5C%20mol%20%5C%20FeCl_3%7D%7B162.19%20%7D)
Divide.
![2.12713484 \ mol \ FeCl_3](https://tex.z-dn.net/?f=2.12713484%20%5C%20mol%20%5C%20FeCl_3)
3. Round
The original measurement of grams, 345.0, has 4 significant figures. We must round our answer to 4 sig figs.
For the answer we calculated, that is the thousandth place.
The 1 in the ten thousandth place tells us to leave the 7 in the thousandth place.
![\approx 2.127 \ mol \ FeCl_3](https://tex.z-dn.net/?f=%5Capprox%202.127%20%5C%20mol%20%5C%20FeCl_3)
There are about <u>2.127 mole</u>s of iron (III) chloride in 345.0 grams.