Scientific Measurements        2

Measurements in Science        2

Temperature        3

Derived Units        4

Significant Figures        4

Precision in Measurement – Significant Figures        4

Significant Figure Rules        4

Multiplication & Division with Sig Figs        5

Adding and Subtracting with Sig Figs        5

Sig Figs with Log Values        5

Scientific Notation        9

Dimensional Analysis        10

Units In Measurements and Problem Solving        10

Concentration        11

Molarity        11

Molality        11

Percent by Mass        11

Example: Concentration Unit Conversion        11

Making solutions        13

Example: Making Solutions 1        13

Example: Making Solutions 2        13

The Equilibrium Constant        14

Manipulating the Equilibrium Expression        14

For example: Given the following information,        15

This Section of the course is largely a review of Gen Chem concepts to be sure we are on the same page. Some of this is disjointed - this is how Harris presents it too.

Scientific Measurements

Measurements in Science

Numbers are useless without units. Reporting you have 27.4 doesn't mean anything. 27.4 what? T-Rexes? Park benches? Molecules? Units complete the answer.

We use the International System of Units (SI) for the majority of our calculations in chemistry.

Mass and weight are used interchangeably in a lot of situations. Strictly speaking, mass is the amount of matter in an object or sample. Weight is mass in a gravitational field. Since we take all of our masses on Earth and our gravitational field is fairly constant, you'll hear me say "mass" and "weight" interchangeably. Weight would be different if we were to measure our matter on the Moon. The mass would be the same.

Become familiar with the prefixes that change the base unit. For example:

1 kilogram = 1000 grams

1 milliliter = 10-3 liters  

Be able to convert between a number of units in this way.

Temperature

When reporting a temperature it is important to note the scale on the thermometer. Most of our labs require Celsius thermometers. You will be responsible for understanding how to convert between each of the common units of thermal energy.

Kelvin = Celsius temperature + 273.15K <= some books teach this as 273 K that works for me too

Fahrenheit = (9F/5C) * (temperature in Celsius) + 32F

The Kelvin scale is considered absolute. Absolute zero, 0 K, is a theoretical temperature. There was a team of researchers who claimed to achieve lower than 0 K but there has been an issue with reproducibility.

Derived Units

Volume and density are common but derived units. That means they depend on a basic unit for their existence. Volume in Liters or milliliters (mL is more common in here) is derived from a length measurement. A cubic centimeter is the same volume as a milliliter (mL).

Density is a ratio of mass to volume. Mass is a basic measurement and volume is derived. Density is derived from the basic mass and length measurements that volume comes from.

Density = mass / volume = g/mL = g/cm3

Significant Figures

Precision in Measurement – Significant Figures

There will be a lot of calculations in this class. You will need to be able to tell the difference between exact and inexact numbers. Precision is the reproducibility of a measurement. Accuracy is how close the measured value is to a true value.

An exact number is a defined value. There are 12 eggs in 1 dozen. Both 12 and 1 are exact. There are 2.54 cm in 1 inch. Both 2.54 and 1 are exact numbers. These are defined.

An inexact number is a number that typically comes from a measured quantity. There is an error in the measurement. There are unavoidable differences in how we see and interpret these measurements.

In science we display the confidence and precision in our measurements with Significant Figures.  Significant figures (Sig Figs) are the meaningful digits in a reported value.

Significant Figure Rules

1. All nonzero digits are significant.
2. Interior zeros (zeros between two numbers) are significant.
3. Trailing zeros (zeros to the right of a nonzero number) that fall after a decimal point are significant.
4. Trailing zeros that fall before a decimal point are significant
5. Leading zeros (zeros to the left of the first nonzero number) are NOT significant. They only serve to locate the decimal point.
6. Trailing zeros at the end of a number, but before an implied decimal point, are ambiguous and should be avoided.

Significant digits within calculations have two primary rules. For multiplication and division, the measured quantity with the least significant digits determines how many significant digits are in the answer.

Multiplication & Division with Sig Figs

The number with the least significant figures, 22.80, determines the number of significant figures in the answer.

Adding and Subtracting with Sig Figs

For adding and subtracting, the precision is determined by the number of digits beyond the decimal. A measurement read to the thousandths place is more precise than a measurement to the tenths place, therefore the answer, when those numbers are subtracted, the answer can only be as precise as the tenths place. In the example above, the hundredths place is less precise than the ten thousandth’s place so the answer can only be to the hundredths place.

Remember the standard math rules when working problems with multiple operations. Keep one digit past the sig figs you need for rounding. Don't round until the end. When observing a measuring instrument, like a ruler, all of the known digits are read and one digit is estimated.

Sig Figs with Log Values

Taking the log of a number is identifying the exponent of 10 that number represents. (I don’t teach math - the words get muddy for me)

n = 10a is also log n = a

a is made up of “characteristic” and “mantissa”

Log 4.5 x 10-6 = -5.3467

-5 is the characteristic; 3467 is the mantissa.

The number of digits in the mantissa of the log should be equal to the number of significant figures of n.

Log 4.5 x 10-6 = -5.3467 ~ -5.35

This is covered in Harris as well, but then later in the book goes on to say when we calculate pH, just round to the hundredth’s place. Since I don’t know where you are going from this course, be sure you understand this concept of characteristic and mantissa.

The last significant digit is the first uncertain digit.

How do you know the first Uncertain digit?

The uncertainty of glassware is often etched or printed on the glassware. The buret on the left has +/- 0.05mL indicating that the buret should be read to the hundredth’s place. It is graduated to the tenth’s place therefore, the hundredth’s place is estimated. The 10mL volumetric flask on the right has +/- 0.025mL printed in very small print next to the 10. When that flask is filled with the bottom of the meniscus level with the printed line on the neck of the flask, the volume is within 0.025mL of 10mL.  These are the uncertainties of the glassware. A balance has uncertainty too.  The uncertainty is usually printed near the display; if it is not, +/- the last display place value is used. (Weegschaal met leeg en gedroogd filter, GNU Free Documentation License.)

The “d=0.0001g” noted on the balance image is the “readability”. This balance will read to the 0.1mg. The “e=0.001g” is the verified or legal trade interval. This balance may measure masses to the 1mg for legal trade.  What about the UNCERTAINTY. There is a method for calculating the uncertainty of a balance, but for now, we’ll use the “readability” as the uncertainty.

(https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Active_Learning/Shorter_Activities/Data_Analysis_and_Statistics/01_Uncertainty)

To report a buret reading, read the volume of the known digits, estimate one more then add the uncertainty. For example, I would record the following image as 22.15 +/- 0.05mL.

Uncertainty and error will propagate as the values are used in calculations. If the buret reading above was the initial volume and the final volume was 41.50 +/- 0.05mL then the difference between the two volume readings would indicate how much volume was delivered.

Volume delivered = Final Volume - Initial Volume

But what happens to the uncertainties?  

They propagate. When adding or subtracting, the uncertainty in the answer will be found by the equation:

R = the answer; A, B, C are the addends or subtrahends.

uA is the uncertainty of the A value; uB is the uncertainty of the B value, uC is the uncertainty of the C value

uR is the uncertainty of the answer.

Volume delivered = Final Volume - Initial Volume

Volume delivered = (41.50 +/- 0.05mL) - (22.15 +/- 0.05mL)

Volume delivered = 19.35 +/- ???mL

uR = square root ( (0.05)2 + (0.05)2 ) = 0.070710678  (?How many digits do we keep?)

The last significant digit is the first uncertain digit.

Errors are typically reported to 1 significant figure.

(I can feel the look on your face.)

The volume delivered was 19.35mL our error will be in the hundredth’s place: 0.07mL. (1 sig fig; hundredth’s place is the first uncertain digit so last significant figure.)

Volume delivered = 19.35 +/- 0.07mL

The 0.07mL is called the standard error of this measurement. We can also report the relative error which is typically reported as the relative PERCENT error:

eR% = 100 * uR / R = 100 * 0.07 / 19.35 = 0.36 % ~ 0.4% (1 sig fig)

If that volume of liquid was massed and had a mass of 20.120 +/- 0.002g, what is the density of that liquid?

Density is mass divided by volume. Multiplication and division of uncertainty is found by:

eR% is the relative percent error.

Density = mass / volume

Density = (20.120 +/- 0.002g) / (19.35 +/- 0.07mL)

Density = 1.03979 +/- ??? g/mL

eA% = 100 * 0.002 / 20.120

eB% = 100 * 0.07 / 19.35

eR% = square root ( (100 * 0.002 / 20.120)2 + (100 * 0.07 / 19.35)2 ) = 0.36189 %

eR% = 100 * uR / R

uR = eR% * R / 100

uR = 0.00376

Density = 1.040g/mL +/- 0.4%

Density = 1.040 +/- 0.004 g/mL

Scientific Notation

When a number is reported in scientific notation, only the significant digits are found in the coefficient (also known as the decimal part). Scientific notation has two parts, the decimal and the exponent. If the exponent is positive, the number is larger than 1. If the exponent is negative, the number is smaller than 1. You will need to become familiar with how your calculator uses Scientific Notation, specifically how YOU enter it. (image attribution: www.boundless.com)

 In scientific notation, the non-zero number is always a number between 1 and 10

1.00 x 10 4 kg = 1.00 x 10,000 = 10,000 kg

2.203 x 10 2 cm = 2.203 x 100 = 220.3 cm

1.00 x 10 - 4 m = 1.00 x 1/10,000 = 0.000100 m

When converting numbers to scientific notation, if the number is greater than one, the exponent increases one unit (from zero) for every place the decimal is moved left to attain the non-zero number between 1 and 10. If the number is less than one, the exponent decreases one unit (from zero) for every place the decimal is moved right to attain the non-zero number between 1 and 10.

0.00000000567 cm is 5.67 x 10-9 cm decimal moved 9 places right

1,230,000,000 m is 1.23 x 109 m decimal moved 9 places left

Using Scientific Notation removes the ambiguity from reporting numbers. A number like 200 (1 significant digit) does not appear to have the same precision as 2.00 x 102 (3 significant digits).  Scientific Notation does not need to be used all the time; only when there is ambiguity to the number of significant digits. A number like 752 is just as precise as 7.52 x 102.  (they both have 3 significant digits) Use your discretion.

Dimensional Analysis

Units In Measurements and Problem Solving

Units are important in calculations and the tracking of units throughout a calculation is essential.The basic SI units were already covered.

A quantitative measurement is one that results in a numerical assessment.  It is something that gives a number to be used in calculations.  In contrast, qualitative, is an observational assessment that could be color or general "feeling" (hot, cold, smooth, rough).  A quantitative measurement needs to have a number and a unit.  In science, we primarily use the metric system (meters, liters, Celsius, grams).  You will need to know how to convert units.

To convert from one unit to another you need to have equality statements.  An equality statement is something like:

100 cm = 1 m

To convert 42.5 inches to meters you would need 2 equality statements.  The centimeter to meter statement above and

1 in = 2.54 cm

Using the algebraic (from algebra) principle that the units will cancel if one is in the numerator and one is in the denominator, the following problem can be set up:

Please notice that the equality statement can be used like a fraction in the conversion problem and the units can be cancelled as if they are factors.  Sometimes this technique is called dimensional analysis or solution mapping.

Concentration

Concentration is the way a solution is described mathematically. A concentration is a representation of how much solute is in a solvent. In Chem 1, molarity was introduced.

Molarity

Molarity = M = moles of solute/ L of solution

Molality

Molality is the number of moles of solute that is dissolved in 1 kg of solvent. This can be calculated for any mass of solvent by using:

Molality = m = moles of solute / mass of solvent (kg)

Keep in mind that molality is m and molarity is M. They are different.

Percent by Mass

Percent by mass or percent by weight is the ratio of masses of a solute to the mass of a solution, multiplied by 100.

Percent mass = mass of solute / mass of solution * 100 %

The mass of the solution is often the mass of solute + mass of solvent.

If we were to multiply the above equation by 1000 rather than 100, we would have the parts per thousand (ppt) concentration; replace 100 with 1,000,000 and we would have parts per million or replace 100 with 1,000,000,000 and we have parts per billion or ppb.

PPM can also be mg/L =  ug/mL

PPB can also be ng/mL =  μg/L

Example: Concentration Unit Conversion

(Glucose with numbering notation.. Provided by: Bermanel. License: CC BY: Attribution.)

You should be able to interconvert between many types of concentration units.

If we have a 0.396 m solution of glucose (C6H12O6) with a density of 1.16 g/mL and we wanted to convert this to molarity and percent by mass we would take the following steps.

Use the definition of molality to determine the mass of glucose in the solution.

0.396 m = 0.396 mole / 1 kg solvent

0.396 mole glucose * (180.2g / 1 mol glucose) = 71.4g C6H12O6

Find the total mass of solution = solvent + solute = 1000g + 71.4g = 1071 g solution

Now calculate the volume of the solution, using the density.

1071 g solution * (1mL/1.16g) * (1L/1000mL) = 0.923 L solution

We know how many moles there were from the definition of molality, 0.396 moles and now we have calculated the volume in Liters so we can calculate the molarity.

M = moles solute / L solution = 0.396 mole / 0.923 L = .429 M

To calculate the percent by mass,

Percent mass = mass of solute / (mass of solute + mass of solvent) * 100 %

Percent mass = 71.4g / (71.4g + 1000g) * 100 % = 6.67%

This solution of glucose can be described as 0.396 m = 0.429M = 6.67%

Making solutions

Volumetric glassware is often used when making solutions to a target concentration. Stock solutions can be purchased in high concentration then diluted as needed. To dilute a stock solution, use the equation

M1V1  =  M2V2

The “M” in the above equation is commonly molarity, but can be any concentration unit. The concentration units do need to be the same on both sides of the equation.

Example: Making Solutions 1

Explain how you would make 100.0mL of a 0.500M NaCl solution.

First, calculate the mass of NaCl needed:

0.1000L NaCl solution * (0.500mol NaCl / 1 L NaCl solution) * (58.44 g NaCl / 1 mol NaCl) = 2.922g NaCl

~ Mass out 2.92g NaCl and quantitatively transfer the solids to a 100.0mL volumetric flask. Fill to the mark with deionized water, cap, and mix.

Example: Making Solutions 2

What volume of 37% HCl (w/w%) is needed to make 250.0mL of 0.100M HCl? Density of 37% HCl is 1.2 g/mL at 25 °C. (https://www.sigmaaldrich.com/US/en/product/sigald/320331)

First - what is 37% in molarity?

If there is 1L of solution, what mass is solute?

1L of solution has a mass of (1000mL * 1.2g/mL = 1200g)

37% = 100 * mass solute / mass solution

37% = 100 * mass solute / 1200g

Mass of solute in 1L of 37% HCl is 444g HCl

Convert to moles and then molarity:

444g HCl * (1 mol HCl / 36.46g HCl) = 12.1777 mole HCl

12.1777 mole HCl / 1 L HCl solution = 12.1777 M HCl

Then - Use the dilution equation OR dimensional analysis

Dilution Equation: M1V1 = M2V2

12.177M * V = 0.100M * 250.0mL  <=  I can use mL if there is an understanding that the answer will also be mL. The molarity units will cancel out.

V = 2.05 mL of stock solution is needed to make 250.0mL of 0.100M HCl

Dimensional analysis method:

0.2500L dilute * (0.100 mol HCl / 1L dilute solution) * (1L concentrated / 12.177 mol HCl) = 0.00205 L concentrated solution

Either method yields the same answer. It doesn’t matter to me which one you choose as long as you consistently get the correct answer.

The Equilibrium Constant

Equilibrium is a state of a reaction system where the forward and reverse reactions occur at the same rate and the concentrations of the reactants and products are not changing. At first, reactants could be introduced where they would produce products. Then the products would react to produce reactants. When this system "evens out" it is at equilibrium.  An equilibrium can be established by introducing a number of different combinations of reactants and products.  Equilibrium is a dynamic state - both the forward and reverse reactions continue to occur but there is no net change in the concentrations over time. Chemically this is shown with an arrow going both ways between the reactants and products. *Since the products, as written in the forward direction, are also the reactants of the reverse, it is accepted that as the reaction is written, the chemicals on the left are the reactants. The chemicals on the right are the products.

aA + bB ⇆ cC + dD

An equilibrium can be described by an equilibrium constant. This is based on the reaction as it is written in the forward direction. This constant is related to the concentration of all aqueous or gaseous species. (no solids and no pure liquids end up in the equilibrium constant expression)

The same expression can be used for the reaction quotient (Qc) at any point in the reaction. At equilibrium, it is called Keq.  Solids and pure liquids do not end up in the equilibrium expression. Because of that, we use molarity to describe liquid solutions and we use bars to describe the pressure of gases.

Manipulating the Equilibrium Expression

The equilibrium expression can be manipulated in a number of ways to suit the reaction system that is to be studied. A reaction can be written in the reverse direction which would cause the Kc value to be the reciprocal value. For example:

2NO(g)  + O2(g) ⇆ 2NO2(g)

If the equation is reversed, 2NO2(g) ⇆ 2NO(g)  + O2(g)

If the equation is multiplied by 2, the Kc equation becomes squared and so does the value of Kc.

4NO(g)  + 2O2(g) ⇆ 4NO2(g)

These manipulation techniques can be used, similar to using Hess's Law with thermodynamics, to calculate steps of reactions and the resulting K value.

For example: Given the following information,

HF(aq) ⇆ H+(aq) + F-(aq)              Kc1=6.8x10-4 (at 25 oC)  (Equation 1)

H2C2O4(aq) ⇆ 2H+(aq) + C2O42-(aq)    Kc2 = 3.8x10-6 (at 25 oC)  (Equation 2)

determine the value of Kc for the following reaction at 25 oC:

C2O42-(aq) + 2HF(aq) ⇆ 2F-(aq)  + H2C2O4(aq)   (Target Equation)

First thing I see is that the first reactant in the target equation (C2O42-(aq)) is a product of Equation 2. I will flip the reaction around which means the new Kc will be the reciprocal.

2H+(aq) + C2O42-(aq)  ⇆ H2C2O4(aq)  Kc2' = 1/Kc2 = 1/3.8x10-6 = 2.6x105

The second reactant of the Target Equation is 2 times the reactant of Equation 1. Multiplying Equation 1 by 2, squares the value of Kc.

2HF(aq) ⇆ 2H+(aq) + 2F-(aq)              Kc1'' = Kc12 = (6.8x10-4)2 = 4.6x10-7

Combining these equations:

2H+(aq) + C2O42-(aq)  ⇆ H2C2O4(aq)                   Kc2' = 2.6x105

2HF(aq) ⇆ 2H+(aq) + 2F-(aq)              Kc1'= 4.6x10-7

2H+(aq) + C2O42-(aq) +  2HF(aq)  ⇆ 2H+(aq) + 2F-(aq) + H2C2O4(aq)   Kctarget = Kc2' * Kc1'

Cancel like terms:

2H+(aq) + C2O42-(aq) +  2HF(aq)  ⇆ 2H+(aq) + 2F-(aq) + H2C2O4(aq)   Kctarget = Kc2' * Kc1'

The chemicals are now the same as the target equation and Kctarget can be calculated.

C2O42-(aq) + 2HF(aq) ⇆ 2F-(aq)  + H2C2O4(aq)  

Kctarget = Kc2' * Kc1' = (2.6x105) * (4.6x10-7) = 0.1196 ~ 0.12