Question

Assume you recovered 0.120 mL of citral in this experiment. How many grams of lemon grass oil did you begin with if citral comprises 5.72% of the oil?

Answer 1

1.84 g/mL

Step-by-step explanation:

We know that

percentage of citral =
`(Volume of citral)/(Volume of sample )`

⇒
`(5.72)/(100) =(0.120)/(V_(sample))`

solving for V_sample we get 2.09 mL

Now density of the lemon grass oil = 0.88g/mL at room temp.

then mass in gm= Volume×Density= 2.09×0.88 =1.84 gram.

Answer 2

1.88 g

Step-by-step explanation:

Considering:-

`\%\ of\ the\ citral=(Volume_(citral))/(Volume_(Sample))* 100`

Given that:-

% Citral = 5.72 %

Volume of the citral = 0.120 mL

So,

`5.72=(0.120\ mL)/(Volume_(Sample))* 100`

`Volume_(Sample)=(0.120* 100)/(5.72)\ mL`

Volume of the grass oil sample = 2.10 mL

Also, Density of the lemon grass oil = 0.893 g/mL

Mass = Density*Volume = 0.893 g/mL*2.10 mL = 1.88 g

1.88 g of lemon grass oil did you begin with if citral comprises 5.72% of the oil.