Archive for February, 2008

Calculating Empirical Formula of Urea When The Masses of Its Components Are Known

Chemistry Tutor Site Exercise
Urea one of useful fertilizer contains 1.121 g N, 0.161 g H, 0.480 g C, and 0.640 g O. What is the empirical formula of urea?

Chemistry Tutor Site Solution
Because the mass of each element in urea has been known, so we can calculate the moles of the atoms:

Mole N = 1.121 / 14.01 = 0.08 mol

Mole H = 0.161 / 1.008 = 0.1597 mol

Mole C = 0.480 / 12.01 = 0.039 mol

Mole O = 0.640 / 16.00 = 0.04 mol

Next step is dividing each mole value by 0.039 ( the smallest number of moles present) we obtain:

N : H : C : O = (0.08/0.039) : (0.1597/0.059) : (0.039/0.039) : (0.04/0.039)

N : H : C : O = 2 : 4 : 1 : 1

So the empirical formula of urea is N2H4CO or we can write as (NH2)2CO

 

Add comment February 17th, 2008

Calculating Molar Mass and Amount of Molecules of Aspirin When The Formula and Weight is known

Chemistry Tutor Site Exercise

The molecular formula of acetylsalicylic acid (aspirin), one of the most commonly used pain relievers, is C9H8O4.

  1. calculate the molar mass of aspirin
  2. a typical aspirin tablet contains 500 mg of C9H8O4. How many moles of C9H8O4 molecules and how many molecules of acetylsalicylic acid are in a 500 mg tablet?

Chemistry Tutor Site Solution

The molar mass is the mass in grams of one mole of the compound. The molar mass is obtained by summing the masses of the component atoms. In one mole of aspirin there are 9 moles of carbon atoms, 8 moles of hydrogen atoms, and 4 moles of oxygen atoms.

9 C: 9x 12.01 = 108.09 g

8 H: 8 x 1.008 = 8.064 g

4 O: 4 x 16.00 = 64 g

Mass of 1 mole C9H8O4 = 132.154 g

So the molar mass of aspirin is 132.154 g

Moles of aspirin are determined by calculate the ratio between its mass and the molar mass. The mass of 1 mole of aspirin is 132.154 g, thus 5 x 10-2 g ( 500 mg) is much less than 1 mole, you can calculate using this fraction:

5 x 10-1 g x ( 1 mol / 132.154 g ) = 3.78 x 10-3 mol of aspirin

Since 1 mole is 6.022 x 1023 units, we can determine the number of molecules:

3.78 x 10-3 x 6.022 x 1023 molecules = 22.76×1020 molecules

and can be rounded to be 23×1020 molecules

 

 

Add comment February 16th, 2008

Calculating Mass Percent of Each Element in Acrylic Acid C3H4O2

Chemistry Tutor Site Exercise
Acrylic acid wich formula C3H4O2 is important starting material for synthetic of acrylic plastic. Calculate the mass percent of each element in acrylic acid.

Chemistry Tutor Site Solution

The masses of the elements in 1 mole of Acrylic acid are

Mass of C = 3 x 12.01 = 36.03 g

Mass of H = 4 x 1.008 = 4.032 g

Mass of O = 2 x 16.00= 32.00 g

Mass of 1 mol of acrylic acid = 72.062 g

Next we find the fraction of the total mass contributed by each element and convert it to a percentage:

Mas percent of C = (36.03 / 72.062) x 100% = 49.99%

Mas percent of H = (4.032 / 72.062) x 100% = 5.60 %

Mas percent of O = (32.00 / 72.062) x 100% = 44.41 %

Check:
Sum the individual mass percent of each element, they should total of 100% within round. In this case the percentages add up to 100.00%

Add comment February 16th, 2008


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