Percent Composition and Standard Deviation calculations

The Mathematics behind the Molecular Weight, Percent

Composition and Standard Deviation Calculations

(Concept from Molar Mass program by Christopher King at Eastern Oregon State College).

For Addition:

In general: For f = x + y - z, the standard deviation of f, s_f, is equal to Sqrt[(s_x)² + (s_y)² + (s_z)²], which is in words, the std. dev. of f is equal to the square root of the sum of the squares of the std. deviations of each term in the original sum.

Applied to the Molecular Weight of a molecule:

MW = ax + by + cz + ..., where a is the number of atoms of weight x, b is the number of atoms of weight y, etc. The Std. Dev. of MW, s_mw = Sqrt[a(s_x)² + b(s_y)² + ...]

NOTE: The reason the equation uses a(s_x)² is that if the terms were added up consecutively 'a' times for the same element, we would have (s_x)² + (s_x)² + (s_x)² etc. up to 'a' times, which combines to be a(s_x)²

For example, for chlorobenzene, C6H5Cl:

MW = 6*12.011 + 5*1.00794 + 35.4527 = 112.5584

s_mw = Sqrt[6*(.001)² + 5*(.00007)² + (.0009)²] = .002614

Resulting MW is 112.558 (±0.003) = 112.558(±3)

For Multiplication:

In general: For f = x*y/z, the standard deviation of f, s_f is equal to f*Sqrt[(s_x / x)² + (s_y / y)² + (s_z / z)²], which is in words, the std. dev. of f is equal to f times the square root of the sum of squares of the quotients of the std. deviations of each term divided by each term itself.

Applied to the Percent Composition of a molecule:

%Comp = (a*x/MW)*100, where a is the number of atoms of weight x and MW is the total molecular weight of the molecule. The Std. Dev. of %Comp, s_%c = %c*Sqrt[(s_a/a)² + (s_x/x)² + (s_mw/mw)²]. But, since a is a finite number, its s_a = 0, so the formula becomes: s_%c = %c*Sqrt[(s_x / x)² + (s_mw / mw)²].

For example, for chlorobenzene, C6H5Cl:

%Comp of Carbon = 6*12.011/112.5584*100 = 64.02543%

s_%c = 64.02543*Sqrt[(.001/12.011)² + (.002614/112.5584)²] = .00553

Resulting %Comp of Carbon = 64.025% (±0.006) = 64.025(±6)%

%Comp of Hydrogen = 5*1.00794/112.5584*100 = 4.47741%

s_%c = 4.47741*Sqrt[(.00007/1.00794)² + (.002614/112.5584)²] = .0003279

Resulting %Comp of Hydrogen = 4.4774% (±0.0003) = 4.4774(±3)%

%Comp of Chlorine = 35.4527/112.5584*100 = 31.49716%

s_%c = 31.49716*Sqrt[(.0009/35.4527)² + (.002614/112.5584)²] = .0010837

Resulting %Comp of Hydrogen = 31.497% (±0.001) = 31.497(±1)%

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