Question

4) Analysis of a sample of nitroglycerine yielded 0.6053 g of carbon, 0.0841 g of hydro 0.7061 8 of nitrogen and 2.4210 g of oxygen. The empirical formula of nitroglyce Solution: 5) People with severe contact dermatitis may have some concern over the nickercom stainless steel utensils. Suppose you knew that the % Ni in a steel sample was approximately 10.00%. If you wanted to ensure that you produced 5.00 gram precipitate, how many grams of steel would you suggest as a sample? Given Ni reacts with a precipitating agent called DMG in a 1:2 stoichiometry to form NI(DMG). The Molar mass of Ni = 58.69 g/mol; the molar mass of Ni(DMG)2 = 288.91 g/mol. Solution:

Answer 1

4)

we have mass of each elements as:

C: 0.6053 g

H: 0.0841 g

N: 0.7061 g

O: 2.421 g

Divide by molar mass to get number of moles of each:

C: 0.6053/12.01 = 0.0504

H: 0.0841/1.008 = 0.0834

N: 0.7061/14.01 = 0.0504

O: 2.421/16.0 = 0.1513

Divide by smallest:

C: 0.0504/0.0504 = 1

H: 0.0834/0.0504 = 5/3

N: 0.0504/0.0504 = 1

O: 0.1513/0.0504 = 3

Multiply by 3 to get simplest whole number ratio:

C: 1*3 = 3

H: 5/3*3 = 5

N: 1*3 = 3

O: 3*3 = 9

So empirical formula is:C3H5N3O9

Answer: C3H5N3O9

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