1. When ethyl bromide reacts with aqueous KOH, it undergoes hydrolysis.The lone pair on O in OH attracts the hydrogen on the beta carbon( the carbon next to the carbon which has the bromide group) and the potasssium cation attracts the bromide group. The product is ethene and the by products are water and KBr.
2. Amounts of CO2 and H2O in moles are:
"n(CO_2)=\\frac{4.4g}{44g\/mol}=0.1mol"
"n(H_2O)=\\frac{2.7g}{18g\/mol}=0.15mol"
The mass of O2 can also be determined directly since the total mass of reactants is always equal to the total mass of products:
"m(O_2)=4.4+2.7-3.1=4.0g"
The amount of O2 in moles:
"n(O_2)=\\frac{4.0g}{32g\/mol}=0.125mol"
Therefore, the O2 : CO2 : H2O molar ratio is 0.125 : 0.1 : 0.15, which equals 5 : 4 : 6 as the lowest whole number ratio.
Now the overall equation can be written:
nCxHyOz + 5O2 --> 4CO2 + 6H2O
For the equation to become balanced, there should be 4 carbon atoms, 12 hydrogen atoms, and 4 oxygen atoms on the left side, in other words, x : y : z = 4 : 12 : 4. The lowest whole number ratio of this (1 : 3 : 1) determines the empirical formula, which is CH3O. To calculate the molecular formula, the molar masses of the empirical formula unit and the unknown compound should be compared:
"M(CH_3O)=12+3\\times1+16=31g\/mol" , which is 2 times less than the actual molar mass (62 g/mol). Therefore, the subscripts of the empirical formula should be multiplied by 2 giving C2H6O2.
Answer:
Empirical formula CH3O
Molecular formula C2H6O2
3. A prussian blue colouration indicates the presence of nitrogen in the compound. If nitrogen is present in the compound, the sodium extract would contain sodium cyanide formed during fusion. On adding the required reagents, sodium cyanide reacts to form ferric-ferrocyanide which has prussian blue colour.
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