Basic Biological Chemistry Chm-094 Labs for the online course
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In this lab handout in the Problems to solve section there are many examples how this procedure is
followed and how to use the dimensional analysis technique to solve problems.
Table 1. Unit equations
Length 1 in = 2.54 cm (exactly)
1 mi = 1.609 km = 1609 m
1 mi = 5280 ft
1 yd = 3 ft
1 ft = 12 in -10
1 Å (angstrom) = 10 m (exactly)
Pressure 1 mm mercury = 1 torr
1 mm mercury = 13.6 mm water
1 atm = 101325 Pa
1 atm = 1013.250 mbar
1 atm = 760 mm mercury (exactly)
Volume 1 mL = 1 cm 3
1 L = 1 dm3 = 1000 cm3
1 gal = 3.785 L
1 gal = 4 qt = 8 pt = 16 cups
1 tablespoon= 14.79 mL
1 teaspoon = 4.93 mL
1 cup = 237 mL
Radiation dose
equivalent
1 Sv (Sievert) = 100 rem -1
1 Sv = 1.00 J/kg = 1.00 J kg
Area Unit equations are not given. The area of a rectangle can be
calculated by multiplying the length
measurements (same units).
Energy 1 cal = 4.184 J (Joules)
1 Cal = 1000 cal = 1 kcal
(note: 1 Cal is not the same as 1 cal).
Time 1 min = 60 s (or 1 min = 60 sec) 1 hr = 60 min
Frequency 1 Hz (Hertz) = 1/s
Mass 1 lb = 0.4536 kg
1 lb = 16 oz
1 ton (short) = 2000 lb
Temperature o
K (Kelvin) = C + 273.15 o o F = ( C×1.8) + 32
o o F 32 C =
1.8
Table 2. Metric system prefixes and unit equations
Prefix Symbol Numerical value
Numerical value in the exponential form
Example of a unit equation
giga- G 1,000,000,000 109 1 GHz = 109 Hz
mega- M 1,000,000 106 1 MJ = 106 J
kilo- k 1,000 103 1 kg = 1000 g
hecto- h 100 2 10
1 hPa = 100 Pa
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deka- (deca-) da 10 101 1 dam = 10 m
1 100
deci- d 0.1 10-1 1 dm = 0.1 m
centi- c 0.01 10-2 1 cm = 0.01 m
milli- m 0.001 10-3 1 mg = 10-3 g
micro- ? or mcg 0.000001 10-6 1 ?L = 10-6 L or 1 mcL = 10-6 L nano- n 0.000000001 10-9 1 ns = 10-9 s
Table 3. Significant figures rules for math operations and unit conversions.
When calculations are completed, to decide how many sig figs are in the answer, use the rules below.
I. Multiplying and dividing: the result should
have as many significant figures
as the measured number with the fewest significant figures.
II. Adding and subtracting: the result should have as many decimal places as the measured number with the smallest number of decimal places.
III. Exact numbers in conversion factors: ignore sig figs in exact numbers in conversion factors, when deciding how may sig figs the calculated number should have and follow previous two sig fig rules in math operations.
Also, in temperature conversions all the numbers on the formulas are exact (except for the temperature itself).
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IV. Sig figs in temperature conversions: the
only number with sig figs is the original temperature measurement; all other numbers are exact ones – do not use them when figuring out the number of sig figs in the calculated temperature.
V. Converting units within the same
measuring system (metric to metric or English
to English):
ignore sig figs in conversion factors when converting between units from the same system and follow sig fig rules in math operations.
VI. Converting units from different measuring systems (metric to English or English to metric):
for each conversion factor use only a numerator or denominator value that has more significant figures; then follow sig fig rules in math operations.
VII. An exception to the previous rule: when
converting between inches and centimeters ignore sig figs in conversion factors.
VIII. Significant figures and intermediate
calculation steps: do not round off numbers
until the very last step. Rounding off too early
may introduce an error in the result. If asked to report the result from an intermediate step, report it with the correct number of sig figs, but for further steps use the unrounded number.
This rule applies to all the multiple step problems that involve numbers. Keep this in mind throughout the course.
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IX. Significant figures in medicine
dosagerelated calculations:
in this type of problems in this course, the concept
of significant figures is not used. The result of
calculations should be rounded, but not to the
correct number of significant figures, but to the
number of digits determined by the measuring or
dispensing device. Students who will take a medicine-related course later, will be given guidelines how to round off their calculated numbers.
Example 1. 1mL syringes have graduations of 0.01
to 0.99, so the medicine volumes less than 1 mL
should be rounded to two digits after the decimal
point.
Example 2. If a syringe or a measuring cup (2.5
mL, 5 mL, 10 mL) has graduations of 0.1 mL, the
medicine volumes should be rounded to one
digit after the decimal point.
Example 3. If a measuring cup has
graduations of 1 mL, the medicine volumes
should be rounded to 0.5 mL.
Example 4. The number of sig figs for infusion
pump rate calculations will depend on the pump
model.
If the infusion pump rate can be set with the precision of two digits after the decimal point, then the rate should be calculated with this precision in mind. For example, if the rate was calculated to be 25.25278 mL/hr, it should be rounded to 25.25 mL/hr. Rounding the result to 25.3 will lead to an overdose and rounding to 25 will lead to an underdose.
Table 4. Few more things to keep in mind (applies to all labs and problems in this course where
calculations are involved).
a. When using dimensional analysis, treat units like numbers: when the same unit
appears in the denominator and the numerator it cancels out; when two identical units are multiplied, the result is the squared unit.
60 ?????? hr ×
5 hr ? = = 300 min 1 ???? hr
5 m × 5 m = 25 m2
b. When adding or subtracting two measurements make sure that they have the same units (the most common examples include area, volume and temperature calculations).
5.0 m + 7 dm this operation cannot be performed directly.
7 dm first should be converted to m:
7 dm = 0.7 m. Then: 5.0 m + 0.7 m = 5.7 m
c. When done with the units do not forget to perform calculations with numbers.
?? hr × ???? min = = 300 5 hr ? min hr 1 hr
1
d. The negative power next to the unit = . unit
g
2 , 5.1 g cm-3 = 5.1 3 2 min-1 = cm
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e. The same unit can be written in several different forms. Learn to recognize them.
g 5 grams per cubic centimeter = 5 3 = 5 g/cm3 =
cm g = 5 g cm-3 = 5 gcm?3 = 5 g/cc = 5
cc
65 mph = 65 miles per hour = 65 miles/hour mi
= 65 mi hr -1 = 65 hr
f. Standard number format, scientific notation
and calculator format.
On the right there are examples of how one should and should not write the answers (report calculated numbers).
For example, the following operation was
0.06381.22. performed:
The calculator may generate and answer that
looks like 5.23E-3 or 5.23E-3 or 5.23-3.
Please do not copy those numbers into a data
sheet in either of those formats. Be able to
recognize the number behind the calculator
format and write it as
0.00523 (standard format) or
5.23?10-3 (scientific notation). This applies to all labs: answers written in the form
like 5.23E-3 or 5.23E-3 or 5.23-3 will not be
accepted. The last number is not even 0.00523, it
is 0.00699 ( ).
5.
÷
e. Decimal format and time measurements. Keep in mind that 2 hours 30 min = 2.5 hours, and 2 hours 15 minutes is not 2.15 hours
Problems to solve
Each b) problem is worth 7.5 points. To get credit for the problem, show all the work.
Show the work in provided space right under the question and write your final answers on the
blank lines so that your instructor can easily see them.
The answers should be written with the correct number of significant figures.
1a. Atmospheric aerosols are liquid or solid particles that are present in the air. These particles are small and many of them affect human health. On one summer day diameters of fine
aerosol particles in the ambient air in New York City ranged from 56 nanometers to 2.0
micrometers. Convert 56 nm to centimeters and to inches and write your answers in a
common format and in scientific notation.
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How to solve the problem.
* We need to convert nanometers (nm) to centimeters (cm) first.
* We are not dealing with compound units here.
* Given unit: nm, desired unit: cm. These are length unit in the metric system. Metric prefixes are
involved (nano-, centi-) meaning that we will likely need Table 2.
* Again, converting nm ? cm.
You are not expected to know how many nanometers are in 1 centimeter, but you are expected to
know how nanometers and meters are related and how centimeters and meters are related.
Make a two-step conversion: nm ? m ? cm. You may want to use Table 2 if you do not remember the metric prefixes yet: 1 nm = 10-9 m and 1 cm = 0.01 m.
10?9 m 1 cm
56 nm ? 1 nm ? 10 ?2 m = 0.0000056 cm = 5.6 ?10-6 cm
* Next we need to convert 56 nm to inches. From
Table 1: 1 inch = 2.54 cm.
56 nm ? 101 ?nm9 m ? 10 1? cm2 m ? 2. in = ?
= ? in
* Make sure that all answers are written with the correct number of sig figs.
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1b. Convert 2.0 ?m to centimeters and to inches and write your answers in a common
format and in scientific notation.
Write answers using a common format Write answers using scientific notation
2a. The mass of an object is 65.3 kg. The length of the object is 3 yards 1.5 feet. Calculate
the mass of the object in pounds and length in inches.
How to solve the problem.
* These are two separate conversions: one is for the length, another one is for the mass. All
conversion factors can be found in Table 1.
Note that here 3 in 3 yards should be treated as an exact number when figuring out the
number of sig figs.
3 ft
* 3 yd + 1.5 ft = 3 yd ? + 1.5 ft = 9 ft + 1.5 ft = 10.5 ft. 1 yd
? in 126 in 10.5 ft
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lb * 65.3 kg ?
0.
* Make sure that all answers are written with the correct number of sig figs.
Object mass = __143__ lb
Object length = __126__ in.
2b. A patients mass and height are 175 lb and 5 feet 6.5 inches respectively.
Calculate the patients mass in kilograms and height in centimeters. Show
all the work below.
Patients mass = _____________ kg
Patients s height = ___________ cm.
3a. Many gases (including those for medical uses) are sold in pressurized gas cylinders. Those
cylinders come in different sizes. How much gas can fit in a cylinder depends on the cylinder
size and the pressure of the gas inside the filled cylinder.
A company sells oxygen in D-size cylinders and claims that the cylinder holds 350 L of
the gas. Convert this volume of oxygen to mL and to gallons.
How to solve the problem
* 350 L ? mL = 350,000 mL (another way: 350 L ?
1000 mL = 350,000 mL) 0. L 1 L
* gal 92 gal
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* Make sure that all answers are written with the correct number of sig figs.
350 L = 350,000 mL 92 gal
3b. Another company claims that their D-size cylinders hold 425 L of the gas (obviously their cylinders have a higher gas pressure inside when full).
Convert this volume of oxygen to mL and to gallons.
Write both numbers in a standard format and in scientific notation.
425 L = ______________mL ______________ gal
4a. A recipe is asking for 2.5 dL of milk. How many cups is it?
How to solve the problem.
* No diagram is needed here.
* We need to convert dL (deciliters) to cups.
* We are not dealing with compound units here.
* Given unit: dL, desired unit: cup. These are volume units (will need Table 1) and a metric prefix is
involved (deci-) meaning that we will likely need Table 2.
* Again, converting dL? L .
Looking at the Table 1 (the Volume section): L ? gal ? cups.
Important note: we do not need to go through gal ? qt ? pt ? cups, use 1 gal = 16 cups directly.
Putting this information together: dL ? L ? gal ? cups.
Since deci- means 0.1 (Table 2), the needed unit equation is 1 dL = 0.1 L Other
unit equations are 1 gal = 3.785 L and 1 gal = 16 cups.
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* Keep multiplying the original measurement by the conversion factors. Do not write the result of each
conversion step (this may result in an error) – just convert through. Round off at the very last step.
2.5 dL ? 0. L ? gal ? 16 cups = 2.5 × 0.1 × 1 × 16 cups = 1.056803 cup ? 1.1 cup
dL 3. L 1 gal 1 × 3.785 × 1
* Note that the final number was rounded to 2 significant figures. Why to two? Recall the rules of sig figs
when dividing/multiplying, when converting within the same unit system and between different systems.
Let us look at the same conversion sequence but now in terms of sig figs.
cups
inverted the conversion factor).
The number looks reasonable for a cooking recipe. If we have divided by 16 instead of multiplying by
16 the answer would be 0.0041 cup. Have you ever seen a recipe that is asking for 4 thousandth of
a cup of anything?
4b. You decided to make a cake using a recipe from a European newspaper (the cake looked amazing and you just had to try to make it). The recipe calls for 310 g of all-purpose flour.
You looked up the cups-to-ounces conversions and found out that 1 cup contains 4.41
oz of all-purpose flour.
How many cups of flour you should take to make the cake? Note: here the cups and ounces are the mass units, not the volume units.
Where to start: you are given one unit equation already 4.41 oz = 1 cup. This
will be your last conversion since you need cups.
Write down the measurement that you are given:
310 g ? Look up the rest of the conversion factors in Table 1.
* The final answer is in cups (if it was, say, in gal 2
that would mean that we have erroneously
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The mass of flour = ____________ cups
5a. Problems in this group are about the caloric value of various food components. Look at the table below.
Calories per gram kilocalories per gram calories per gram
Proteins 4.0 4.0 4.0?103
Carbohydrates (carbs) 4.0 4.0 4.0?103
Lipids (fats) 9.0 9.0 9.0?103
Note that
* Calorie and calorie are not the same unit (1 Calorie = 1 kcal = 1000 cal – see Table 1);
* although dietary fiber belongs to the carbohydrate group and is included in the total carbohydrate
count on the label, its caloric (nutritional) value is zero (human body does not digest it).
A hamburger contains 13 grams of protein, 8 grams of fat and 31 grams of
carbohydrates. The mass of fiber is 1 g. Calculate the nutritional value of the hamburger
in kilocalories (common format), calories (scientific notation) and Joules (scientific
notation).
How to solve the problem.
4.0 kcal
* From protein: 13 g protein ? = 52 kcal 1 g protein
9.0 kcal
* From fat: 8 g fat ? = 72 kcal 1 g fat
4.0 kcal
* From carbohydrates: (31-1) g carbohydrates ? = 120 kcal 1 g carbohydrates
These three steps are intermediate steps, the answers were not rounded off.
* The nutritional value of the burger is 52 kcal + 72 kcal + 120 kcal = 244 kcal
1000 cal
* The nutritional value of the burger is 244 kcal ? = 244,000 = 2.44?105 cal 1 kcal
4. J
* The nutritional value of the burger is 2.44?105 cal ? = 1,020,896 J ? ? J cal
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5b. A taco contains 8 grams of protein, 9 grams of fat and 13 grams of carbohydrates. The mass of fiber is 3 g. Calculate the nutritional value of the taco in kilocalories (common
format), calories (scientific notation) and Joules (scientific notation).
The nutritional value of the taco is ____________ kcal = ______________ cal = _______________ Joules 6a. You are asked to measure out approximately 45 mL of a dextrose solution.
How would you measure this volume out if the solution comes in a bottle and you only
have a tablespoon and a plastic cup without graduation marks?
How to solve the problem.
* Remember a safe lab practice: never dispense a chemical from its original container.
Pour it in a smaller container a cup if perfect for that.
Then you will need a teaspoon to measure out approximately 45 mL.
* From Table 1: 1 tablespoon = 14.79 mL
Tbsp
* 45 mL ? 3.0 Tbsp * Take three tablespoons of the solution from a plastic cup.
* Note that can only do this because it is specified that the measurement is approximate. If there were
no word approximate, a tablespoon should not be used as a measuring out substances, since it is
not a precise device.
A more precise instrument like a graduated cylinder should be used.
6b. This is a theoretical question do not perform actual measurements (calculations only please).
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You are asked to measure out approximately 10 mL of a salt solution.
How would you measure this volume out if the solution comes in a bottle and you only
have a teaspoon?
How would you measure this volume out if the solution comes in a drop bottle and you
have no other measuring devices? 1 mL contains approximately 20 drops.
10 mL __________ teaspoons
10 mL __________ drops
7a. Ferritin is a blood protein that contains iron. A ferritin blood test is performed to determine how much iron a patients body stores. It is important to know that different laboratories may follow
slightly different guidelines of what to consider the normal range of ferritin. For this problem
let us consider the following ranges of ferritin as normal: 24-336 mcg/L for men and 11-307
mcg/L for women (https://www.mayoclinic.org/tests-procedures/ferritin-test/about/pac-20384928).
A patient got bloodwork done, and her ferritin level was found to be 17 ng/mL. Is
the ferritin concentration within the normal range? If it is not, is this concentration
low or high? Confirm your answer with calculations. Show all the work.
How to solve the problem.
* Before checking whether the ferritin concentration (17 ng/mL) falls within the normal range you should
check the units of the reported concentration (ng/mL which stands for nanograms per milliliter) and
the units for the referenced normal range (mcg/L which stands for micrograms per liter). These do
not look like the same units, so you cannot compare the numbers directly.
First you will have to convert ng/mL to mcg/L.
ng *
When working with a compound unit, do not use a / symbol, rewrite the unit in a ratio format: .
mL ng
17 mL
* Next we will need to convert ng to mcg and mL to L in one line.
https://www.mayoclinic.org/tests-procedures/ferritin-test/about/pac-20384928
https://www.mayoclinic.org/tests-procedures/ferritin-test/about/pac-20384928
https://www.mayoclinic.org/tests-procedures/ferritin-test/about/pac-20384928
https://www.mayoclinic.org/tests-procedures/ferritin-test/about/pac-20384928
https://www.mayoclinic.org/tests-procedures/ferritin-test/about/pac-20384928
https://www.mayoclinic.org/tests-procedures/ferritin-test/about/pac-20384928
https://www.mayoclinic.org/tests-procedures/ferritin-test/about/pac-20384928
https://www.mayoclinic.org/tests-procedures/ferritin-test/about/pac-20384928
https://www.mayoclinic.org/tests-procedures/ferritin-test/about/pac-20384928
Basic Biological Chemistry Chm-094 Labs for the online course
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You are not expected to know how many nanograms are in 1 microgram, so make a two-step
conversion: nanograms (ng) ? grams (g) ? micrograms (mcg). You may want to use Table 2 if you
do not remember the metric prefixes yet: 1 ng = 10-9 g and 1 mcg = 10-6 g. ng 10?9 g 1 mcg
17 mL ? 1 ng ? 10?6 g
* We are not done yet. You need to add a conversion factor mL ? L: ng
10?9 g 1 mcg 1 mL
17 mL ? 1 ng ? 10?6 g ? 10?3 L
Notice that all the conversion factors were written in way that unwanted units cancel out and target
units (mcg and L) stay.
?9 × mcg
* Now perform math operations: = 17 L this is the patients ferritin level.
By performing the conversion, we have shown that 17 ng/mL = 17 mcg/L.
Since a patient is a female this concentration is within the normal range (11 – 307 mcg/L) 17
mcg/L is higher than 11 mcg/L, but lower than 307 mcg/L.
7b. One of the waste products produced by kidneys is creatinine. A healthy range of creatinine in human blood is between 0.50 mg/dL and 1.1 mg/dL. Elevated levels of this
compound could indicate a problem with kidney function.
The concentration of creatinine in patients blood was reported in g/L: 0.0082 g/L. Is
this value within the normal range? If it is not, is this concentration low or high?
Confirm your answer with your calculations and show all the work.
The concentration of creatinine = _______________mg/dL. It is ___________the normal range.
8a. A recommended dose of a medicine is 0.0650 g daily for a 155-lb person.
An actual patient however weighs 185 lb.
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What is the appropriate medication dose for this patient in milligrams (mg)?
* First calculate how much medicine is required per 1 lb of body mass:
??.???????? ?? g ?????? ???? lb
= 0.00041935 (do not round off the answer yet). Note that converting grams to pounds or pounds to gram in attempt to cancel the mass units
here is incorrect here since those masses refer to different things: grams of the medication
and pounds of the human body mass. Do not convert lb to mg here! 0.0650 g medication
0.00041935 g medication
= . 155 lb body mass 1 lb body mass
* Next multiply the body mass of the actual patient by how much medication is needed per
pound of body mass:
0.00041935 g medication
185 lb body mass × = 0.077581 g medication 1 lb body mass Note
which units got cancelled out.
* Round off the answer to the correct number of significant figures and convert the result to mg
(as asked).
1 mg medication
0.077581 g medication 0.0776 g medication × = 77.6 mg 0.001 g medication
* Check if your answer makes sense. It does, since a heavier person will need more medication
than a lighter one.
8b. A recommended dose of a medicine A is 125 mg for a 150 lb person. What would be a correct dose for a 142 lb person?
The only measurement here is 142 lb. Treat all other numbers as exact numbers here.
The appropriate medication dose for this patient is ________________ mg
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9a. A patient needs a dose of 0.75 g of medicine A. He/she was already given 250 mg.
How many more mg does the patient need?
The medicine comes in tablets. Each tablet contains 125 mg of the medicine.
How many tablets did the patient already get?
How many more tablets should the patient get?
How to solve the problem.
* We are asked about mg and the number of tablets. Therefore, we need to start by converting
0.75 g to mg.
mg
0.75 g × = 750 mg 0. g
* The patient needs 750 mg of the medicine, and was already given 250 mg.
The patient needs 500 mg more (750 mg 250 mg). *
If one tablet contains 125 g of the medicine,
1 tablet
500 mg corresponds to 500 mg × = 4 tablets this is what the patient still needs; 125 mg
1 tablet
250 mg corresponds to 250 mg × = 2 tablets this is what the patient already got. 125 mg
9b. Medicine B comes in 30 mg tablets. A nurse is informed that a patient is to receive 180 mg of
B per day, given in three doses.
How many tablets are required per day?
How many tablets are required in each dose for this patient?
To answer the first question, you may want to use the formula:
Total dosage needed = Number of tablets needed ? Dosage per tablet
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The patient needs ______________________ tablets per day.
The patient needs ______________________ tablets per dose.
10a. A liquid medicine contains a certain dose within a given (reference) volume of the liquid, so quantities calculation may require more steps. To calculate the dosage needed for a patient,
it is recommended that you use the following formula:
?????? ???????? ???????????? The volume of the liquid needed = ? The given volume.
?????? ???????? ???? ?????? ?????????? ????????????
Both doses have to be in the same units, both in mg, or both in mcg.
A patient needs a dose of 250 mg of medicine C. C is available in a solution of 400 mg
per 200 mL. How many mL of the solution are needed to produce the needed dose?
How to solve the problem.
* Draw a diagram, even if you think you do not need it this will help you to understand
what you are calculating.
Use the formula suggested above:
250 mg 400 mg
The volume of the liquid needed = ? 200 mL = 125 mL * Note that we did not use the sig fig rules here (see Table 1, Row IX).
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* The last thing to do is to check that your result makes sense. It does: 400 mg in 200 mL.
250 mg is slightly more than a half of 400 mg, so it makes sense that 125
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