Chemistry 121 Fall 2004

homework problems: 50, 52, 56, 58, 60, 62, 64 Note that this is different from what I said in lecture, I was looking at the wrong problem set.

When we use an instrument to take measurements, we have to consider the error involved in the measurement. Neither we nor our instruments are perfect, so there will always be some small error involved --no matter how good our technique is. We use two terms when we talk about error - precision and accuracy.

Figure 1.9

When measuring numbers, the last digit always includes error. So when I use a ruler to measure the width of my paper, then the last digit has the uncertainty. We call these significant figures. Any other figures beyond this last digit are meaningless. Usually these meaningless numbers appear during calculations, so we have to follow some rules when doing calculations to prevent having meaningless numbers in our calculations. The rules are straight forward and listed in section 1.4. I expect you to read this section and learn these rules. You will be using them in the first laboratory and your answers on quizzes, labs and exams will include points for appropriate use of significant figures.

Rules for determining sig figs:

1 . All nonzero digits are significant.

2. Zeros between significant digits are significant.

108, 1.08 – both zeros significant

3. A lone zero preceding a decimal point is not significant.

0.246 – this zero is not significant

4. Zeros to the right of the decimal point that precedes the first nonzero digit is not significant.

0.0098 – non of these zeros are significant

5. Zeros at the end of a number are significant is they are to the right of the decimal.

1.2300 – these zeros are significant

Rules for addition/subtraction

When adding or subtracting, it is the number of decimal places we are concerned with. The number with the fewest decimal places determines how many decimals places the final answer has.

2.25 + 81.3 – 0.668 = 82.9

Since 81.3 has only one decimal place, the answer must have only one decimal place.

Rules for multiplication/division

The result should have no more significant figures than the number with the fewest significant figures.

3.14 x 8.6631 = 27.2

Since 3.14 has only 3 sig figs. then the answer should only have t

Density

density = mass/volume

This can also be used as a conversion between mass and volume. Note that for a particular solution we can relate mass to volume (mass/volume) or volume to mass (volume/mass).

Consider diet coke or regular coke. One floats, the other doesn't. Why? How can we scientifically support our answer?

a. we can determine the density

b. we can change the density of the one that floats

An example problem using density:

Diamond and cubic zirconia cannot be told apart with the naked eye (unless there are inclusions). But they have different densities:

diamond = 3.513 g/cm³

cubic zirconia = 5.68 g/cm³ (ZnO₂)

Gemstones are sold by the carat, a measure of mass. 1 carat = 0.20 grams exactly

Thought question: Which is bigger, a one carat diamond or a one carat cubic zironia? Consider the density and how it is relating mass and volume. The bigger the density, the less volume it takes up for a particular mass.

Cubic Zirconia is made by melting ZnO₂, adding colorants then recrystallizing it. If we recrystallized 2.50 g of ZnO₂ into a cube, what would the edge of the cube measure in mm?

volume = 2.50 g x 1 cm³/5.68g = 0.440 cm³

note the we have 3 sig figs in each initial value so the answer must have 3 sig figs.

volume = length x width x height. For a cube they are all the same so:

volume = (edge length)³ or edge length = (volume)^0.333 (a way of taking the cube root)

edge length = (0.440 cm³)^0.333 = 0.761 cm x 10 mm/ 1cm = 7.61 mm

How many carats would a similarly sized diamond end up being?

Since the size is the same, then volume is the same.

mass in carats = 0.440 cm³ x 3.513g/cm³ x 1 carat/0.20g = 7.73 carats

(A really big diamond!)