1.2 Unit Conversions

Density example

An 18.2 g sample of zinc metal has a volume of 2.55 cm3 .Calculate the density of zinc. Step 1: List the known quantities and plan the problem. Known mass = 18.2 g volume = 2.55 cm3 Unknown density = ? g/cm3 Use the equation for density, D = m/V, to solve the problem. Step 2: Calculate. D = m V = 18.2 g 2.55 cm3 = 7.14 g/cm3 Step 3: Think about your result. If 1 cm3 of zinc has a mass of about 7 grams, then a sample that is approximately 2.5 cm3 will have a mass between 2 and 3 times its density, which is consistent with the values given in this problem. Additionally, metals are expected to have a density that is greater than that of water, and zinc's density falls within the range of the other metals listed above ( Table 1.4). Since density values are known for many substances, density can be used to determine an unknown mass or an unknown volume. Dimensional analysis will be used to ensure that units cancel appropriately.

Derived Unit conversion example

Convert 3.6 mm3 to mL. Step 1: List the known quantities and plan the problem. Known • 1 m = 1000 mm 1 mL = 1 cm3 1 m = 100 cm Unknown 3.6 mm3 = ? mL This problem requires multiple steps and the technique for converting with derived units. Simply proceed one step at a time: mm3 to m3 to cm3 = mL. Step 2: Calculate. 3.6 mm3 × 1 m 1000 mm3 × 100 cm 1 m 3 × 1 mL 1 cm3 = 0.0036 mL Numerically, the steps are to divide 3.6 by 109 , followed by multiplying by 106 . Alternatively, you can shorten the calculation by one step if you first determine the conversion factor between mm and cm. Using the fact that there are 10 mm in 1 cm, you can avoid the intermediate step of converting to meters. 3.6 mm3 × 1 cm 10 mm3 × 1 mL 1 cm3 = 0.0036 mL Both conversion methods give the same result of 0.0036 mL. Step 3: Think about your result. Cubic millimeters are much smaller than cubic centimeters, so the final answer is much less than the original number of mm3

Density

Is the ratio of the mass of an object to its volume. Density = mass/volume or D = m/V It is an intensive property, so it doesn't depend on the amount of material present in the sample. Ex. water has a density of 1.0 g/L Its SI units are kg/m^3 because kg and m are the SI units for mass and length. Although most liquids and solids have densities that are conveniently expressed in g/cm^3. Since a cubic centimeter is equal to a millimeter density can also be expressed as g/L. Gases are much less, so their densities are normally written in g/L. Since most materials expand as temp increases the density of a substance is temperature dependent and decreases as the temp increases.

Density of common substances

Liquids and Solids Density at 20°C (g/mL) Ethanol 0.79 Ice (0°C) 0.917 Corn oil 0.922 Water 0.998 Water (4°C) 1.000 Corn syrup 1.36 Aluminum 2.70 Copper 8.92 Lead 11.35 Mercury 13.6 Gold 19.3

Two step metric converison

Some metric conversion problems are most easily solved by breaking them down into more than one step. When both the given unit and the desired unit have prefixes, one can first convert to the simple (unprefixed) unit, followed by a conversion to the desired unit. Two-Step Metric Conversion:Convert 4.3 cm to µm. Step 1: List the known quantities and plan the problem. Known: 1 m = 100 cm 1 m = 10^6 µm Unknown: 4.3 cm = ? µm You may need to consult the table in the lesson, "The International System of Units," for the multiplication factor represented by each metric prefix. First convert cm to m; then convert m to µm. Step 2: Calculate. 4.3 cm× 1 m 100 cm × 106 µm 1 m = 43,000 µm Each conversion factor is written so that unit of the denominator cancels with the unit of the numerator of the previous factor. Step 3: Think about your result. A micrometer is a smaller unit of length than a centimeter, so the answer in micrometers is larger than the number of centimeters given.

How do you solve a dimensional analysis problem?

Step 1: List the known and unknown qualities and plan the problem. Step 2: Calculate Step 3: Think about your result Example: How many seconds are in a day? Known: 1 day=24 hour 1 hour=60 min 1 min=60 sec Unknown: 1 day= ? sec (the known qualities represent the conversion factors that you will use) The first conversion factor will be a day in the denominator, so it cancels out, the second one will be in hours, so hours cancels out, and so on (the unit of the last numerator will be the units used for the answer since everything else has been canceled out) 1 d x 24h/1 d x 60 min/1 h x 60 s/1 min= 86,400 s (in applying the first conversion factor, the "d" cancels out and 1 x 24=24, in the second, the "h" cancels out, so 24 x 60=1440, and in the third, the "min" cancels out so 1440 x 60=86400 (step 3: the second is a smaller unit than day or hours, so it makes sense that there is a larger amount of numbers of seconds in one day)

using density to determine mass and volume

What is the mass of 2.49 cm3 of aluminum? What is the volume of 50.0 g of aluminum? Step 1: List the known quantities and plan the problem. Known density = 2.70 g/cm3 1. volume = 2.49 cm3 2. mass = 50.0 g Unknown 1. mass = ? g 2. volume = ? cm3 Use the equation for density, D = m/V, and dimensional analysis to solve each problem. Step 2: Calculate. 2.49 cm3 × 2.70 g 1 cm3 = 6.72 g 50.0 g× 1 cm3 2.70 g = 18.5 cm3 In problem one, the mass is equal to the density multiplied by the volume. In problem two, the volume is equal to the mass divided by the density. Step 3: Think about your results. Because a 1 cm3 sample of aluminum has a mass of 2.70 g, the mass of a 2.49 cm3 sample should be greater than 2.70 g. The mass of a 50-g block of aluminum is substantially more than the value of its density in g/cm3 , so that amount should occupy a volume that is significantly larger than 1 cm3 .

Using dimensional analysis with derived units

When units are squared or cubed, as with area or volume, the conversion factors themselves must also be squared. Shown below is the conversion factor for cubic centimeters and cubic meters. (1 m/100 cm)^3= 1 m^3/10^6 cm^3 = 1 Because a cube has 3 sides, each side is subject to the conversion of 1 m to 100 cm. Since 100 cubed is equal to 1 million (10^6 ), there are 10^6 cm^3 in 1 m^3. Two convenient volume units are the liter, which is equal to a cubic decimeter, and the milliliter, which is equal to a cubic centimeter. The conversion factor would be: (1 dm/10 cm)^3 = 1 dm^3/1000 cm^3 = 1 There are thus 1000 cm^3 in 1 dm^3 , which is the same thing as saying there are 1000 mL in 1 L.

Metric Unit Coversions

example:a particular experiment requires 120 mL of a solution. The teacher knows that he will need to make enough solution for 40 experiments to be performed throughout the day. How many liters of solution should he prepare? Step 1: List all known and unknown quantities Known: 1 experiment requires 120 mL of solution 1 L=1000 mL Unknown: 40 experiments require? L of solution Since each experiment requires 120 mL of solution and the teacher needs to prepare enough for 40 experiments multiply 120 by 40 to get 4800 mL of solution needed. Now you must convert mL to L by using a conversion factor. Step 2: Calculate 4800 mL x 1 L/1000 mL=4.8 L Note that conversion factor is arranged so that the mL is in the denominator and this canciles out leaving L as the remaining unit in the answer. Step 3: Think about your result A liter is much larger than a milliliter, so it makes sense that the number of liters required is less that the number of milliliters.

Conversion Factor

is a ratio of equivalent measurements For example: both 1 m and 100 cm can represent the same length because the value of the conversion factor is 1 (it is read as 1 m per 100 cm) The numerator and denominator all must represent equal quantities in each case to be valid conversion factor (otherwise equal 1) used in solving problems where a certain measurement must be expressed with different units, when a given measurement is multiplied by an appropriate conversion factor the numerical value changes but the actual size of the quantity measured stays the same

Dimensional Analysis

is a technique that uses the units (dimensions) of the measurement in order to correctly solve problems

Derived Unit

is a unit that results from a mathematical combination of SI base units. Include: Area, Volume, Density, Concentration, Speed (velocity), Acceleration, Force, Energy

What is useful about dimensional analysis?

it is useful to convert from one metric system unit to another

What is useful about the metric system?

it's many prefixes allow for quantities to be expressed in many different units

Acceleration

quantity: Acceleration symbol: a unit: meters per second squared unit abbreviation: m/s^2 derivation: speed/time

Area

quantity: Area symbol: A unit: square meter unit abbreviation: m^2 derivation: length x width

Concentration

quantity: Concentration symbol: c unit: moles per liter unit abbreviation: mol/L derivation: amount/volume

Density

quantity: Density symbol: D unit: kilograms per cubic meter unit abbreviation: kg/m^3 derivation: mass/volume

Energy

quantity: Energy symbol: E unit: joule unit abbreviation: J derivation: force x length

Force

quantity: Force symbol: F unit: newton unit abbreviation: N derivation: mass x acceleration

Speed (velocity)

quantity: Speed (velocity) symbol: v unit: meters per second unit abbreviation: m/s derivation: length/time

Volume

quantity: Volume symbol: V unit: cubic meter unit abbreviation: m^3 derivation: length x width x height