Dimensional analysis (also called factor label method or unit analysis) is used to convert from one set of units to another.  This method is used for both simple (feet to inches) and complex (g/cm³ to kg / gallon) conversions and uses relationships or conversion factors between different sets of units.  While the terms are frequently used interchangeably, conversion factors and relationships are different.  Conversion factors are quantities that are equal to one another, such as 100 cm = 1m, 100 cm = 1m, in which both values describe a length.  Relationships are between two values that are not necessarily a measure of the same quantity.  For example, the density of water is 1.00 g / mL.  Grams are a measure of mass while milliliters measure volume so this is considered a relationship rather than a conversion factor.  Either way, we depend on units to help set up and solve the calculation.  We will see additional examples of relationships as we explore other details about chemical substance.

Conversion Factors

Many quantities can be expressed in several different ways.  The English system measurement of 4 cups is also equal to 2 pints, 1 quart, and 1414of a gallon.  Notice that the numerical component of each quantity is different, while the actual amount of material that it represents is the same.  That is because the units are different.  We can establish the same set of equalities for the metric system: 

1meter = 10 decimeters = 100 centimeters = 1000 millimeters 1 meter = 10 decimeters = 100 centimeters = 1000 millimeters.

The metric system’s use of powers of 10 for all conversions makes this quite simple.  We can write conversion factors between any pair of equivalent quantities.  In each conversion factor, the numerator and denominator represent equal quantities so they are all valid conversion factors. Additionally, these conversion factors can be inverted or used in combination with other conversion factors in a dimensional analysis problem.

1 meter = 10 decimeters = 100 centimeters = 1000 milimeters

Case1: Conversion factor can be used within a system of units or between different units of system. (defined relationship)

Example problem: A lab test showed an individual’s cholesterol level to be 186 mg/dL. Convert this quantity into g/dL

given 1g= 1000mg

In order to obtain metric to metric conversion factor, the meaning of metric prefixes and their values must be known.

For more information watch:

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Metric conversion: ladder method

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Figure 1.21: Ladder method of Metric System

Ref: commons.wikimedia.org/

How do you use ladder method?

1st- Determine your starting point.

2nd- Count the “jumps” to your ending point.

3rd- Move the decimal the same number of jumps in the same direction.

Example:  4km  =__________m

                    ↑start                     ↑End

4.           0                  0              0 = 4000m

Write the abbreviation of following Metric Conversions:

1) Kilogram _____                 4) Milliliter _____                  7) Kilometer _____

2) Meter _____                       5) Millimeter _____                8) Centimeter _____

3) Gram _____                        6) Liter _____                         9) Milligram _____

Try these conversions, using the ladder method.

10) 2000 mg = _______ g      15) 5 L = _______ mL           20) 16 cm = _______ mm

11) 104 km = _______ m       16) 198 g = _______ kg         21) 2500 m = _______ km

12) 480 cm = _____ m           17) 75 mL = _____ L             22) 65 g = _____ mg

13) 5.6 kg = _____ g             18) 50 cm = _____ m             23) 6.3 cm = _____ mm

14) 8 mm = _____ cm            19) 5.6 m = _____ cm            24) 120 mg = _____ g

Compare using <, >, or =.

56 cm         6 m                                  

7 g          698 mg

5 g         508 mg         

1,500 mL         1.5 L

536 cm          53.6 dm            

43 mg         5 g                         

3.6 m         36 cm

For more information visit:

https://www.metric-conversions.org/

Questions:

1. Convert each of the following measurements into meters?

1. 2.5 × 10³ mm
2. 25 μm

Case 2: Metric to English conversion factors are specified with different number of significant figures. (measured relationship). They are called equalities.  For example: 1 mile= 1.61 km. mile is the English unit and kilometer (km) in the metric unit.

Example #1 : Capillaries, the microscopic vessels that carry the blood from small arteries to small veins, are on the average only 0.1cm long. What is the average length of a capillary in inches?

Given 1 inch= 2.54 cm

0.1 cm × 1 inch         =   0.0393 in= 0.04 in ( lowest 1 sig fig used in calc)

                2.54 cm

Here is the table of most common metric to metric and metric to English conversion factors:

Example #2: Convert 10.8 ft to meters

Given that  12 in = 1ft,  2.54 cm= 1 inch and 100 cm = 1meter

Pathway: ft * 12 in             *  2.54 cm                  *  1 m         = meter
                          1 ft                    1 in                        100 cm

Now plug in the numbers in tis problem:

10.8 ft * 12 in    * 2.54 cm  * 1m = 3.28 m ( ans)
                 1ft *   1 in *    100cm

Example#3:  Convert 58.8 lbs to kg.

Given : 58.8 lbs,  needed  kg. conversion factor: 2.205 lbs= 1 kg

pathway:  lbs *   1kg

205. lbs

now use the values given in the problem,

58.8 lbs *  1 kg  = 26.7 kg

                2.205

Table 1.6 Unit Conversion factors between Metric & English

For more information watch:

For more information visit:

https://learner.org/interactives/metric/conversion_chart.html

Practice Questions:

1. A chemistry student weighs 83.1 kg and 1.93 m tall. What are the person’s equivalent measurements in pounds and feet?
2. What volume of water in gallons would be required to fill a 30.ml of container?

Example 1

How many centimeters are in 3.4m?

Solution

This problem requires the conversion from one unit to another so we can use dimensional analysis to solve the problem.  We need to identify the units that are given (m), the units for the answer (cm), and any relationships that relate the units of the known and unknown values.  In this case, we will use the relationship of 1 m = 100 cm. Start with the known value and its unit.

3.4 m× ? = ___ cm

Then, we look at the units of our relationship to see which value goes in the numerator and which value goes in the denominator.  Remember, we are trying to find the value in centimeters. Since our known value is in units of meters, we need meters to be in the denominator so that it will cancel.  As a result, centimeters will be in the numerator.

3.4 m × 100 cm / 1 m

Note that the numbers stay with the appropriate unit (100 with centimeters and 1 with meters). Now, the meters will cancel out and we are left with units of centimeters.  Always check that your problem is set up completely and that your units cancel correctly before you do the actual calculation.

3.4 m × 100 cm / 1 m = 340 cm = 3.4 × 10² cm

We find the answer to be 340 cm or 3.4 × 10² cm.

Derived Units

Using dimensional analysis with derived units requires special care.  When units are squared or cubed as with area or volume, the conversion factors themselves must also be squared or cubed. Two convenient volume units are the liter, which is equal to a cubic decimeter, and the milliliters, which is equal to a cubic centimeter.  There are thus 1000 cm³ in 1dm³, which is the same thing as saying there are 1000 mL in 1L. The conversion factor of 1cm³ = 1 mL is a very useful conversion.

Example 2

Convert 3.6 mm³ to mL.

Determine the units of the known value, mm³ and the units of the unknown value, mL. The starting and ending units will help guide the setup of the problem.  Next, list any known conversion factors that might be helpful.

1 m = 1000 mm

1 mL = 1cm³

1 m = 100 cm

Now, we can set up the problem to find the value in units of mL.  Once we know the starting units, we can then use the conversion factors to find the answer.

3.6 mm³ × (??)

Continue to use the conversion factors between the units to set up the rest of the problem.  Note that all of the units cancel except mL, which are the requested units for the answer.  Since the values in these conversion factors are exact numbers, they will not affect the number of significant figures in the answer.  Only the original value (3.6) will be considered in determining significant figures.

3.6 mm³ × (1m / 1000 mm)³ × (??)

Once you have solved the problem, always ask if the answer seems reasonable.  Remember, a millimeter is very small and a cubic millimeter is also very small.  Therefore, we would expect a small volume which means 0.0036mL0.0036mL is reasonable.

If you find that you forgot to cube numbers as well as units, you can setup the problem in an expanded form which is the equivalent to the previous method to cube the numerical values.

Step#1: Identify the units of known or given quantity and the units of target quantity

Step#2: Figure out the conversion factors required from given unit to target unit.

Step#3: Multiply the given quantity by one or more conversion factors in such a manner that the original units are canceled leaving only the desired unit.

Watch this out to solve drug dosage related problems that require conversion factors.

Example Problem:

1. A dose of 250 mg of acetaminophen is prescribed for a 20-kg child. How many ml of Children’s Tynenol (100. mg of acetaminophen per 2.5 ml) are needed?

Step#1: Identify given and target quantity and numbers:

Given: 250mg             target: ml

Step#2: Write the conversion factors:

Provided in the problem:                    2.5 ml
                                                            100 mg

Step#3. Multiply the given quantity by the conversion factor.

250 mg ×  2.5 ml      =  6.3ml (lowest 2 sig fig)
                  100 mg

Check: the answer is reasonable. Since the required dose is larger than standard dose, volume would be larger than standard.

Nurses Conversion units for drugs

Drug Calculations

Drug calculations vary depending on whether you are dealing with liquid or solid medications, or if the dose is to be given over a period of time. In this section I will go over each of these situations in turn.

It is very important that you know how drug dosages are worked out, because it is good practise to always check calculations before giving medication, no matter who worked out the original amount. It is far better to point out a mistake on paper than overdose a patient.

a)  Tablets

Working out dosage from tablets is simple.

Formula for dosage:

Total dosage required  = Number of tablets required *Dosage per tablet

Note-If your answer involves small fractions of tablets, it would be more sensible to try to find tablets of a different strength rather than try to make 3  of a tablet for example.

Examples

1. A patient needs 500mg of X per day. X comes in 125mg tablets. How many tablets per day does he need to take?

Total dosage required is 500mg, Dosage per tablet is 125mgSo our calculation is 500/125 = 4. He needs 4 tablets a day

b)  Liquid Medicines

Liquid medicines are a little trickier to deal with as they will contain a certain dose within a certain amount of liquid, such as 250mg in 50ml, for example.

To work out the dosage, we use the formula:

What you want   × What it’s in 

What you’ve got

Note: In order to use this formula, the units of measurement must be the same for “What you want” and “What you’ve got ; i.e. both mg or both mcg etc.

Examples

2. We need a dose of 500mg of Y. Y is available in a solution of 250mg per 50ml.

In this case,

What we want = 500 What we’ve got = 250 What it’s in = 50

So our calculation is 5 0 0 × 50 =100

250

We need 100ml of solution.

3. We need a dose of 250mg of Z. Z is available in a solution of 400mg per 200ml. In this case,

What we want = 250 What we’ve got = 400 What it’s in = 200

So our calculation is 2 5 0 × 200 = 125

400

We need 125ml of solution.

c)  Medicine over Time

1. Tablets/liquids

This differs from the normal calculations in that we have to split our answer for the total dosage into 2 or more smaller doses.

Look at Example 1 again. If the patient needed the 500mg dose to last the day, and tablets were taken four times a day, then our total of 4 tablets would have to be split over 4 doses.

Total amount of liquid/tablets for day = Amount to be given per dose Number of doses per day

We would perform the calculation: 4÷4=1 So he would need 1 tablet 4 times a day.

2)  Drugs delivered via infusion

For calculations involving infusion, we need the following information:

2. • The total dosage required
   • The period of time over which medication is to be given
   • How much medication there is in the solution

Example

4. A patient is receiving 500mg of medicine X over a 20 hour period. X is delivered in a solution of 10mg per 50ml.

What rate should the infusion be set to? Here our total dosage required is 500mg

Period of time is 20 hours

There are 10mg of X per 50ml of solution

Firstly we need to know the total volume of solution that the patient is to receive. Using the formula for liquid dosage we have:

 500/10 × 50 = 2500 So the patient needs to receive 2500mls.We now divide the amount to be given by the time to be taken: 2500/20 =125The patient needs 2500mls to be given at a rate of 125mls per hour.Note: Working out medicines over time can appear daunting, but all you do is work out how much medicine is needed in total, and then divide it by the amount of hours/doses needed.

d)  Drugs labelled as a percentage

Some drugs may be labelled in different ways to those used earlier.

V/V and W/V

Some drugs may have V/V or W/V on the label.

V/V means that the percentage on the bottle corresponds to volume of drug per volume of solution

i.e 15% V/V means for every 100ml of solution, 15ml is the drug.

W/V means that the percentage on the bottle corresponds to the weight of drug per volume of solution.  Normally this is of the form ‘number of grams per number of millilitres’. So in this case 15% W/V means that for every 100ml of solution there are 15 grams of the drug.

If we are converting between solution strengths, such as diluting a 20% solution to make it a 10% solution, we do not need to know whether the solution is V/V or W/V.

Examples

5. We need to make up 1 liter of a 5% solution of A. We have stock solution of 10%. How much of the stock solution do we need? How much water do we need?

We can adapt the formula for liquid medicines here:

What we want      ×  What we want it to be in 

What we’ve got

 We want a 5% solution. This is the same as 5/100 or 1/20.

We’ve got a 10% solution. This is the same as 10/100 or 1/10 .

We want our finished solution to have a volume of 1000ml.

Our formula becomes: (1/20)×1000

                                  (1/10)

(using the rule for dividing fractions)

1/2×1000=500 We need 500mls of the A solution.

Which means we need 1000 – 500 = 500mls of water.

(Alternatively you can use the fact that a 5% solution is half the strength of a 10% solution to see that you need 500ml of solution and 500ml of water)

6. You have a 20% V/V solution of drug F. The patient requires 30ml of the drug. How much of the solution is required?

20% V/V means that for every 100ml of solution we have 20ml of drug F. Using our formula:

What you want   × What it’s in

What you’ve got

This becomes 30/20×100 = 150 We need 150mls of solution.

7. Drug G comes in a W/V solution of  5%. The patient requires 15 grams of G. How many mls of solution are needed?

5% W/V means that for every 100mls of solution, there are 5 grams of G. Using the formula gives us

15/5 × 100 = 300

300mls of solution are required.

Other Dimensional analysis problem example:

2. A person is driving a car with a speed of 229.8 km/h. What is the speed in

1. Miles per hour
2. Feet per second

Given 1 km= 0.6214 mile 1000m= 1km,  1 m= 3.28 ft, 60 sec=1min and 60 min= 1 hr

1. Step#1: Identify given and target quantity and numbers:

Given: 229.8km/hr                 target: mi/hr

Step#2: Write the conversion factors:

Provided in the problem:                    0.6214 mi
                                                            1 km

Step#3. Multiply the given quantity by the conversion factor.

229.8 km ×  0.6214 mi      =  143.4mi/hr (lowest 4 sig fig used in problem)
                  1 km

2. Step#1: Identify given and target quantity and numbers:

Given: 229.8km/hr                 target: ft/s

Step#2: Write the conversion factors:

Provided in the problem:                    3.28 ft & 0.6214 mi
                                                            1 m          1 km

1 hr,              1 min

60 min           60 sec

Step#3. Multiply the given quantity by the conversion factor.

229.8 km ×  1000m  × 3.28 ft   × 1 hr       × 1 min

1 hr             1 km         1 m      60 min      60 sec

=  209 ft/s ( lowest 3 sig fig used in calc).

For more information watch this out:

For more practice visit:

https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/01._Introduction%3A_Matter_and_Measurement/1.6%3A_Dimensional_Analysis

Questions:

1. If one teaspoon contains 5.0 ml, how many teaspoons of Children Tylenol  (100. mg of acetaminophen per 2.5 ml) are needed for a child with a dose of 240 mg?
2. A patient is prescribed 0.150mg of a drug that is available in 25 μg tablets. How many tablets are needed?
3. How many milliliters of Children’s Mortin (100 mg of ibuprofen per 5 ml) are needed to give a child a dose of 180 mg?