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Matter and Math. Chemistry 131 Chapter 1. Matter . Formulas . Physical Properties. Dimensional Analysis. Matter and its Classification. - PowerPoint PPT PresentationTRANSCRIPT
Matter and Math
Chemistry 131Chapter 1
Physical Properties
DimensionalAnalysis
Matter Formulas
Matter and its Classification
This section deals with how we categorize the types of matter into distinct and coherent subgroups. In addition it build a connection map between the different groups.
MatterEverything of substance around you is matter. Whatever has mass is matter. There is a lot more to it than this, but we will start here.
Can be subdivided into…
Pure Substances
&
Mixtures
Matter can be divided into two subgroups, the Pure Substance and the Mixture. Pure substances are those composed of only one element, while mixtures are a combination of at least two elements.
Can be further subdivided into…
Heterogeneous Mixtures
&
SolutionsHeterogeneous mixtures can be physically separated while solutions cannot. Solutions must be chemically separated.
Matter can be divided into two subgroups, the Pure Substance and the Mixture. Pure substances are those composed of only one element, while mixtures are a combination of at least two elements.
Can be further subdivided into…
Pure Substances
&
Mixtures
Compounds
&
ElementsCompounds are where more than one atom is bonded together in some fashion. Elements are individual atoms.
Can be further subdivided into…
Molecular Compounds
&
Ionic CompoundsMolecular compounds, or molecules, are those where the atoms are covalently bonded together. The atoms in ionic compounds are held together by ionic bonds.
Compounds
&
ElementsCompounds are where more than one atom is bonded together in some fashion. Elements are individual atoms.
Can be further subdivided into…
Three Classes, or Groups…
Metals
Metals are those elements on the far right and middle of the periodic table. They hold their outer electrons loosely, have luster, are ductile, malleable, and conductive.
Metalloids
Metalloids have properties of both metals and nonmetals, they have luster are typically ductile or malleable to a limited degree and are semi conductive.
Nonmetals
Nonmetals are brittle in their solid forms (though they may be in gaseous or liquid forms at room temperature). Typically they are electrical insulators.
Now for the big picture…
MatterMixtures
Pure Substances
ElementsMetals
MetalloidsNonmetals
Compounds Ionic Compounds
Molecular Compounds
SolutionsHeterogeneous Mixtures
So how do we describe compounds?
Chemical Formulas
Chemical formulas are the shorthand of chemistry, the language if you will. There are rules to its grammar and syntax that we need to learn and follow as we go through the course.
What is the compound composed of?
Take the molecule apart, down to its constituent atoms.
What is there?
Now you need to identify the atoms by their elements, in our kits they have standard colors, black for carbon, red for oxygen and white for hydrogen.
Oxygen
CarbonHydrogen
How many of each?
Oxygen = 1
Carbon = 2Hydrogen = 6
Can we abbreviated anything?Writing out the names of each element every time would be tedious, so each element has an abbreviation, a chemical symbol, to represent it. These symbols may be obvious (O for oxygen) or obscure (W for tungsten, because it used to be Wolfram). These are the symbols used on the Periodic Table.
Oxygen = O
Carbon = CHydrogen = H
So what do we have now?
Oxygen = 1 OCarbon = 2 CHydrogen = 6 H
Now add the grammar?The grammar, or syntax, of chemistry is such that elements present as a single atom are represented by their symbol. Elements that are represented by more than one atom are have their atomic symbol followed by a subscript with the number of atoms for that element.
Oxygen = OCarbon = C2
Hydrogen = H6
Start with the element with the highest number of bonds.Well get into this more later, but the highest number of bonds means the highest number of available electrons and these are the core of how chemistry works.
Oxygen = OCarbon = C2
Hydrogen = H6Bonds are the holes in this model… little hard to see
Okay… C2… then what?
Add the rest in…C2H5OH
Wait… why not C2H6O2 or C2O2H6?
Sometimes it isn’t as simple…Let us look at the compound again
Notice how the OH can be taken off as a distinct unit?
Functional Groups… hence…C2H5OHFunctional groups are substructures of a compound with specific structure and similar functions. They are written separately in the formulas to emphasize their role in the overall structure and function of the compound.
Now you try one
CH3OH
Physical Properties
This section deal with the difference between what it means to undergo a physical versus chemical change.
What is a Physical Property?
Physical Properties of matter are those that can be measured or observed without the matter undergoing a chemical change.
Observations can be eitherQualitative or Quantitative
Qualitative is Subjective
Observations can be eitherQualitative or Quantitative
Quantitative is Objective
What types of Quantitative measures can we do?
Mass
Mass is a measure of the amount of matter you have. This is recorded in grams (g) or kilograms (kg). Commonly this is referred to as weight, though technically weight is the mass within a specific gravity field. In other words, the weight may vary, but the mass will not.
Volume
Volume is a measure of the amount of space matter occupies. This is commonly measured in units such as liter (L), milliliter (mL) or cubic meter (m3)
Density
Density is a derived measurement, it is calculated by dividing the mass of an object by its volume.
Dimensional Analysis
In this section we are going to work on problems to strengthen your ability to do the basic math required in this course.
Time for the MATH!
Ooo, dreams of fractions
It isn’t that bad, really.
Liar…
No Honest… let’s try…
Try this one.𝑋=( 12 )
2
× 12+14
Answer = 3/8
1. Square 1/2 = 1/4 2. Multiple that by 1/2 = 1/83. Add that to 1/4, with the least common
denominator as 8 = 3/8
How about this one?Solve the following for R.
Answer:
1. divide nT from both side of the equation.
Now a little more… try to figure out these conversions.
Hint, check your books for the conversion factors.
Complete the following conversions
A. 1.54 kg = __________gB. 3.46 cm = __________ µmC. 12.4 ml = __________ lD. 2.3 × 103 in = ___________ m
15403460000.0124
58.42
Try these, they are density equations.
A. A liquid that has a mass of 125 g per 120 ml, what is the density in g/ml?
B. A solid has a mass of 0.025 kg per 100 cm3, what is the density in g/cm3?
1.04 g/ml
0.25 g/cm3
Now to ramp it up a bit…Calculate the density of the following materials in g/ml.A. 49 lbs./fl. oz.B. 3.79 × 10-4 kg/l
A. 751.65 g/mlB. 3.79 × 10-4 g/ml
How to use Dimensional Analysis
Trust me, it will help.
Let us look at this one again…Calculate the density of the following materials in g/ml.A. 49 lbs./fl. oz.
What are you trying to determine?
Let us look at this one again…Calculate the density of the following materials in g/ml.A. 49 lbs./fl. oz.
How many grams/ml?
Right, but let’s rewrite that…
Let us look at this one again…Calculate the density of the following materials in g/ml.A. 49 lbs./fl. oz.
? 𝒈𝒎𝒍=¿
Now put in what we know already…
Let us look at this one again…Calculate the density of the following materials in g/ml.A. 49 lbs./fl. oz.
? 𝒈𝒎𝒍=
𝟒𝟗 𝒍𝒃𝒔𝒇𝒍 .𝒐𝒛 .
Now we need to go from one set of units to the other, do them one at a time. Let’s start with the pounds…
Let us look at this one again…Calculate the density of the following materials in g/ml.A. 49 lbs./fl. oz.
? 𝒈𝒎𝒍=
𝟒𝟗 𝒍𝒃𝒔𝒇𝒍 .𝒐𝒛 . ×
𝟒𝟓𝟑 .𝟔𝒈𝟏.𝟎 𝒍𝒃𝒔 .
Notice how pounds are above and below the fraction line. These cancel out each other. If we solve it here, we would be in g/fl. oz.
Let us look at this one again…Calculate the density of the following materials in g/ml.A. 49 lbs./fl. oz.
? 𝒈𝒎𝒍=
𝟒𝟗 𝒍𝒃𝒔𝒇𝒍 .𝒐𝒛 . ×
𝟒𝟓𝟑 .𝟔𝒈𝟏.𝟎 𝒍𝒃𝒔 .
Now do the same with the fluid ounces to milliliters.
Let us look at this one again…Calculate the density of the following materials in g/ml.A. 49 lbs./fl. oz.
? 𝒈𝒎𝒍=
𝟒𝟗 𝒍𝒃𝒔𝒇𝒍 .𝒐𝒛 . ×
𝟒𝟓𝟑 .𝟔𝒈𝟏.𝟎 𝒍𝒃𝒔 . ×
𝟏 𝒇𝒍 .𝒐𝒛 .𝟐𝟗 .𝟓𝟕𝒎𝒍
See how each step is another step towards what we were looking for, now the fl. oz. cancel.
Let us look at this one again…Calculate the density of the following materials in g/ml.A. 49 lbs./fl. oz.
? 𝒈𝒎𝒍=
𝟒𝟗 𝒍𝒃𝒔𝒇𝒍 .𝒐𝒛 . ×
𝟒𝟓𝟑 .𝟔𝒈𝟏.𝟎 𝒍𝒃𝒔 . ×
𝟏 𝒇𝒍 .𝒐𝒛 .𝟐𝟗 .𝟓𝟕𝒎𝒍
Now when we plug and chug the only units left are g/ml… time to hit the calculator.
Let us look at this one again…Calculate the density of the following materials in g/ml.A. 49 lbs./fl. oz.
? 𝒈𝒎𝒍=
𝟒𝟗 𝒍𝒃𝒔𝒇𝒍 .𝒐𝒛 . ×
𝟒𝟓𝟑 .𝟔𝒈𝟏.𝟎 𝒍𝒃𝒔 . ×
𝟏 𝒇𝒍 .𝒐𝒛 .𝟐𝟗 .𝟓𝟕𝒎𝒍
? 𝒈𝒎𝒍=
𝟒𝟗×𝟒𝟓𝟑 .𝟔×𝟏𝟏×𝟏×𝟐𝟗 .𝟓𝟕 =
𝟐𝟐𝟐𝟐𝟔 .𝟒𝟐𝟗 .𝟓𝟕 =𝟕𝟓𝟏 .𝟔𝟓 𝒈
𝒎𝒍
Significant Figures
You can only be as precise as your least precise measurement.
Not Accurate or Precise.
Precise and Accurate
0
1
2
3
4
We account for precise through significant figures.
The question of zeros
Zeros that denote magnitude are not significant.
Zeros that are part of the measurement are significant
Examples
2305.0 = 5 significant digits
0.00456 = 3 significant digits
52,000,000 = 2-8 significant digits, these cases must be determined by context of the measurement.
Problems1. 345.00 = ______ sig figs 2. 0.0030045 = ______ sig figs 3. 4500.001 = ______ sig figs 4. 4.6 × 106 = ______ sig figs
5
5
7
2
In math, do not consider exact numbers (such as conversion factors)
The answer must have the same number of significant figures as the least precise measured value in the equation.
Examples
2.5 X 3.25 = 8.1 (2 sig figs)
3 X 3.567 = 10 (1 sig fig)
45.3 / 67.9 = 0.667 (3 sig figs)
Problems1. 3.45 × 3.45 = ____________ 2. 4.9/2 = ______________ 3. 3.45×104 × 9.5674 = _______________
11.9
2
3.30×105
Rounding Off AnswersThis is where a lot of students have problems, but the rules are simple.
If the first digit to be dropped is less than 5, leave the last digit kept unchanged.
If the first digit dropped is greater than 5 or is 5 followed by a digit other than zero, raise the last digit kept by one.
If the first digit dropped is 5 followed by only zeros or no other digits, then leave the kept digit unchanged if even and raise by one if odd.
Problems
Given a solution of density 3.87 lbs./gal. how many kilograms of the solution would you have with 1.5 liters of the solution?
0.70 kg
Problems
Nitroglycerin expands at a rate of 1200 times its original volume during detonation, in terms of metric unit volume, how much volume would be generated by a 1.000 pound mass that has a density of 6.870 lbs./ft3?
4947 liters
Problems
Given a solution of density 3.87 lbs./gal. how many kilograms of the solution would you have with 1.5 liters of the solution?
0.70 kg