No, an element is a pure substance, it cannot be broken down into simpler substances or be decomposed by chemical means.
Hello!
We use the amount in grams (mass ratio) based on the composition of the elements, see: (in 100g solution)
C: 70.79% = 70,79 g
H: 8.91% = 8.91 g
N: 4.59% = 4.59 g
O: 15.72% = 15.72 g
Let us use the above mentioned data (in g) and values will be converted to amount of substance (number of moles) by dividing by molecular mass (g / mol) each of the values, lets see:
![C: \dfrac{70.79\:\diagup\!\!\!\!\!g}{12\:\diagup\!\!\!\!\!g/mol} \approx 5.89\:mol](https://tex.z-dn.net/?f=C%3A%20%5Cdfrac%7B70.79%5C%3A%5Cdiagup%5C%21%5C%21%5C%21%5C%21%5C%21g%7D%7B12%5C%3A%5Cdiagup%5C%21%5C%21%5C%21%5C%21%5C%21g%2Fmol%7D%20%5Capprox%205.89%5C%3Amol)
![H: \dfrac{8.91\:\diagup\!\!\!\!\!g}{1\:\diagup\!\!\!\!\!g/mol} = 8.91\:mol](https://tex.z-dn.net/?f=H%3A%20%5Cdfrac%7B8.91%5C%3A%5Cdiagup%5C%21%5C%21%5C%21%5C%21%5C%21g%7D%7B1%5C%3A%5Cdiagup%5C%21%5C%21%5C%21%5C%21%5C%21g%2Fmol%7D%20%3D%208.91%5C%3Amol)
![N: \dfrac{4.59\:\diagup\!\!\!\!\!g}{14\:\diagup\!\!\!\!\!g/mol} \approx 0.328\:mol](https://tex.z-dn.net/?f=N%3A%20%5Cdfrac%7B4.59%5C%3A%5Cdiagup%5C%21%5C%21%5C%21%5C%21%5C%21g%7D%7B14%5C%3A%5Cdiagup%5C%21%5C%21%5C%21%5C%21%5C%21g%2Fmol%7D%20%5Capprox%200.328%5C%3Amol)
![O: \dfrac{15.72\:\diagup\!\!\!\!\!g}{16\:\diagup\!\!\!\!\!g/mol} = 0.9825\:mol](https://tex.z-dn.net/?f=O%3A%20%5Cdfrac%7B15.72%5C%3A%5Cdiagup%5C%21%5C%21%5C%21%5C%21%5C%21g%7D%7B16%5C%3A%5Cdiagup%5C%21%5C%21%5C%21%5C%21%5C%21g%2Fmol%7D%20%3D%200.9825%5C%3Amol)
We note that the values found above are not integers, so let's divide these values by the smallest of them, so that the proportion is not changed, let's see:
![C: \dfrac{5.89}{0.328}\to\:\:\boxed{C\approx 18}](https://tex.z-dn.net/?f=C%3A%20%5Cdfrac%7B5.89%7D%7B0.328%7D%5Cto%5C%3A%5C%3A%5Cboxed%7BC%5Capprox%2018%7D)
![H: \dfrac{8.91}{0.328}\to\:\:\boxed{H\approx 27}](https://tex.z-dn.net/?f=H%3A%20%5Cdfrac%7B8.91%7D%7B0.328%7D%5Cto%5C%3A%5C%3A%5Cboxed%7BH%5Capprox%2027%7D)
![N: \dfrac{0.328}{0.328}\to\:\:\boxed{N\approx 1}](https://tex.z-dn.net/?f=N%3A%20%5Cdfrac%7B0.328%7D%7B0.328%7D%5Cto%5C%3A%5C%3A%5Cboxed%7BN%5Capprox%201%7D)
![O: \dfrac{0.9825}{0.328}\to\:\:\boxed{O\approx 3}](https://tex.z-dn.net/?f=O%3A%20%5Cdfrac%7B0.9825%7D%7B0.328%7D%5Cto%5C%3A%5C%3A%5Cboxed%7BO%5Capprox%203%7D)
T<span>hus, the minimum or empirical formula found for the compound will be:
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I hope this helps. =)
Answer:
0.20 mol's
Explanation:
1.675 L = 1.675 dm^3
moles = V/(conc):
moles = 1.675/(8.5)
moles = 0.1970... --> 0.20
The number of unique open-chain structures corresponding to the molecular formula C₃H₅Cl is two.
<em>Step 1</em>. Draw <em>all possible connections of three carbon atoms</em>.
The is only one possibility (See first image below).
<em>Step 2</em>. Put the <em>Cl in all possible unique positions</em>.
There are only <em>two</em> possibilities: on an end carbon or on the middle carbon (see second image below).
Thus, there are only two isomers of C₃H₅Cl.