[MathisFun]

SOLUTIONS:

--Calculate the number of grams of solid necessary to prepare 50.00 mL of a 0.02000 M solution for the following compounds, C₁₂H₂₂O₁₁ (sucrose) and NaCl (salt).

How many moles of C₁₂H₂₂O₁₁ do I need to do this?
Convert 50.00 ml to 0.05000 L
Using the following equation: M = mol/L, then making the proper adjustment,
mol = M x L = (0.02000 mol/L)(0.05000 L) = 0.001000 mol

Next, how many grams of C₁₂H₂₂O₁₁ equal 0.001000 mol?
Find the molar mass of C₁₂H₂₂O₁₁.
C – 12 – 12.011 – 144.132
H – 22 – 1.00794 – 22.17468
O – 11 – 15.9994 – 175.9934
Which, adds up to: 342.300 g/mol (using 3 digits of precision)

g = (mol)(g/mol) = (0.001000 mol)( 342.300 g/mol) = 0.3423 g (4 significant figures) of sucrose dissolved in water and brought to a final volume of 50.00 mL (0.05000 L) constitutes 0.02000 M solution.

Repeat the procedure for NaCl, except, in this go around, since it is desired to prepare a solution of equal volume and concentration as C₁₂H₂₂O₁₁ in the previous example, I substitute the molar mass of NaCl in the previous equation.
g = (mol)(g/mol) = (0.001000 mol)( 58.4425 g/mol) = 0.05844 g (again, 4 significant figures).

Molar Mass
C₁₂H₂₂O₁₁: 342.300 g/mol
NaCl : 58.4425 g/mol

Mass
C₁₂H₂₂O₁₁: 0.3423 g
NaCl: 0.05844 g

--Calculate how many milliliters of 0.1000 M acetic acid you need in order to prepare 50.00 mL of a 0.02000 M acetic acid solution.

I use the following equation to calculate dilutions: M₁V₁ = M₂V₂, where M1 and V1 serve as the concentration and volume of the original, more concentrated solution, and M2 and V2 are the concentration and volume of the diluted solution.

Convert 50.00 mL to 0.05000 L.

V = ((M₂V₂)/M₁) = ((0.02000 M x 0.05000 L)/0.1000 M) = 0.01000 L (4 significant figures)

Convert 0.01000 L back to mL, 10.00 mL, of 0.1000 M acetic acid added to water and diluted to a total final volume of 50.00 mL will yield 0.02000 M solution.

CH₃COOH (acetic acid)
Stock concentration: 0.1000 M
Volume: 10.00 mL
---
MOLECULAR WEIGHT:

Problems (a) and (b) are provided as examples.

What is the molecular weight of (a) glucose, C₆H₁₂O₆ and (b) urea, (NH₂)₂CO?

(a)   Glucose, C₆H₁₂O₆
C – 6 x 12.0 = 72.0
H - 12 x 1.0 = 12.0
O – 6 x 16.0 = 96.0
C₆H₁₂O₆ = 180.0 amu

(b)   Urea, (NH₂)₂CO
N – 2 x 14.0 = 28.0
H – 4 x 1.0 = 4.0
C - 1 x 12.0 = 12.0
O – 1 x 16.0 = 16.0
(NH₂)₂CO = 60.0 amu

The textbook rounds the figures for these equations. Is this generally accepted, since it’s like giving a rough estimate, much in the same way, if a person were to measure the amount of volume and uses a beaker as opposed to a graduated cylinder?

Here is the actual practice, what is the molecular weight of (c) ibuprofen, C₁₃H₁₈O₂ and (d) formula weight of barium phosphate, Ba₃(PO₄)₂?

My portion will be listen in red text.
(c)   Ibuprofen, C₁₃H₁₈O₂
C – 13 x 12.0 = 156 / 156.0
H – 18 x 1.01 = 18.2 / 18.0
O – 2 x 16.0 = 32.0
C₁₃H₁₈O₂ = 206 amu / 206.0 amu

(d)   Barium Phosphate, Ba₃(PO₄)₂
Ba – 3 x 137 (137.0) = 411 / 411.0
P – 2 x 31.0 = 62.0
O – 8 x 16.0 = 128 / 128.0
Ba₃(PO₄)₂ = 601 amu / 601.0 amu

My approach to problems (c) and (d) follows the same pattern as (a) and (b), by using one digit of precision. My dispute is listed in red text. The parts listed, as having no digit of precision, would be Ba (137 amu), and the sums of ibuprofen (206 amu) and barium phosphate (601 amu). This is the textbook’s answer. I search throughout the chapter, and find no explanation for the ‘arbitrary’ reasoning with using or ‘not using’ digits of precision in this or that case. The only pattern I can draw is that, problems (c) and (d) have amu going into three digits, beyond 99, while problems (a) and (b) are below the value of 100. This set has only caused more confusion for me.

Again, thanks for looking out.

[Dan]

http://en.wikipedia.org/wiki/Significance_arithmetic#Addition_and_subtraction_using_significance_arithmetic

[MathisFun]

Thanks for the link. It's a refresher on the rules. I think my text has a typo.