Chem fundamentals (unit 1)

1.1 The study of chemistry

Matter: physical material; anything with mass that takes up space

Pure substances: matter that has a ==definite composition,== one that does not change, and has distinct properties. They can ==only be separated by chemical reactions==

• Elements: pure substances that cannot be decomposed into simpler substances.
  • Atoms: smallest building block of matter. Each element is composed of 1 type of atom
  • ex: C
  • Molecules: ==2+ atoms,== can be same or different
  • ex: O2, H2O
• Compounds: pure substances ==composed of 2+ different elements==. They can only be separated chemically
  • Ex: H2O

1.2 Classification of Matter

Property: a characteristic to recognize/distinguish types of matter

• Physical properties: can be observed without changing the identity or composition of matter. They are the result of IMFAs between structures (melting point, refractive index, color)
  • Intensive: properties that are ==independent of quantity== (boiling point, odor)
  • Extensive: properties that are ==dependent on quantity== (volume, mass)
• Chemical properties: observed by destroying substance, they result from chemical reactions

Mixtures: a combo of 2+ pure substances. Each substance maintains its own properties. Mixtures can be separated into its pure substances

• Homogeneous: mixtures that are uniform throughout. The components are ==evenly distributed.== They look pure but aren’t since they’re not chemically combined.
  • Solution: small particles, don’t scatter light
  • ex: copper sulfate (aq) or brass
  • Colloid: have large particles, scatter light
  • ex: milk
• Heterogenous: Mixtures that aren’t evenly distributed (granite, wood)
  • Suspension: you can see layers

1.3 Properties of Matter

Physical change: changes physical appearance, not composition

• same substance before and after change
• ex: ice→ water (state change)

Chemical change: substance→different substance

Separation of mixtures

• Distillation: process depends on the boiling points to form gases
  • NaCl + H2O: when you boil it, the water evaporates, leaving salt behind
• Chromatography: depends on the differing size and polarity of substances to adhere to surfaces of solids to separate mixtures
  • Separating chlorophyll pigments in leaves
• Filtration: mix of solids and liquid that’s poured through filter paper. The liquid passes through, leaving the solid in the filter paper
  • ex: coffee

2.1: Atomic theory of matter

Dalton’s atomic theory:

1. Each element is composed of extremely small particles (atoms)

2. All atoms of a given element are identical to each other

   1. All O2 atoms are the same, all N2 atoms are the same

3. Atoms of 1 element can’t be changed into atoms of different elements by chemical reactions.

   1. O2 can’t turn into N2

4. Compounds are formed when atoms of more than 1 element chemically combine.

   1. O + N (elements)→NO (compound)

Law of conservation of mass: matter isn’t created or destroyed, just rearranged

Law of constant composition/definite proportion: g==iven compounds always have same elements in the same proportion.== The ratios are fixed

• H2O is always 2:1 ratio of H:O

Law of multiple proportions: compounds with different ratios of the same atoms are different

• H2O is different from H2O2

2.2 discovery of atomic structure

Democritus: made first atomic model in 400 BC

• proposed that all matter is made up of atoms (small, solid, indivisible particles)
• Model: ball
• 

Dalton: determined that each element is made up of atoms, created atomic theory

• model: ball (same as above image)

Thompson: through cathode ray tube experiments, determined that there are negatively charged electrons

• because electrons contribute a small fractions of atom’s mass, they are small

  

Cathode ray: A glass tube is pumped with air and when high voltage is applied to the electrodes. radiation occurs between them. The radiation (cathode rays, stream of - charged particles) originate at the anode and travel to the cathode. The rays are the same regardless of the cathode material.

He constructed a cathode ray tube w/ a hole in the anode where the cathode ray can pass. Because the electron is negative, the electric field deflects rays in 1 direction. This allowed him to calculate 1.76 x 10^8 Coulombs/g for the ratio of electron charge: mass

Rutherford: through his gold foil experiment, he discovered protons and the nucleus

• Most of the atom’s mass comes from dense + nucleus and most of the volume is empty space (electron cloud)
• Bombarded gold foil with alpha particles---majority passed through with no deflection. Very few particles were deflected-----proved that there’s a small but highly charged nucleus
• 

Chadwick: through nuclear bombardment, he found the neutron

Radioactivity: 3 types of radiation (alpha, beta, gamma)

Beta (-1) are high speed electrons bent towards + end

Alpha (+2) bent towards - end

Gamma (0) are high energy waves, no particles or charge bent towards no end, unaffected

2.3: modern view of atomic structure

Nucleus: contains protons and neutrons with an overall + charge.

• very small and dense (1x10^-15 m)

Electron cloud: contains negatively charged electrons

• almost no mass but most of atom volume (1-5 x10^-10 m)

Angstrom: 1x10^-10m=100pm

Atomic mass unit (amu)=1.66054x10^-24 g

Subatomic particles

Proton: +1, 1.0073amu

Electron: -1, 5.486x10^-4amu

Neutron: 0, 1.0087amu

Isotopic Notation

A: mass number, protons+neutrons

• Isotopes: same # protons but different # neutrons, differing mass numbers

Z: atomic number, just protons, used to identify element

q: charge, #protons vs electrons

2.4: atomic weights

1amu=1.66054x10^-24 g

1g=6.02x10^23 amu

Atomic mass: the number from the periodic table is dependent on isotopic abundance

• Σ (isotope mass)(isotope abundance)

Mass spectrometer:

1. Get atoms into gas phase and convert them into ions (cations)
2. When gas phase cations made, they’re accelerated towards negative grid
3. Only a narrow beam of ions can pass
4. Beam passes through magnet poles that deflect ions
5. Ions separated into their masses (isotopes) ---smaller mass goes first
6. 

Mass spectroscopy: uses spectrometer to determine the mass of an element/molecule

• Provides mass of ions and relative abundance, allows us to calculate atomic mass

• converts atoms/molecules into anions to measure mass and abundance in electric field. The beam of ions passes through poles of a magnet which deflects them on a curved path. The extent of deflection depends on the mass. Ions are sorted according to their mass

  

  Ex: 9/16 atoms have a mass of 70, 6/16 have a mass of 72, 1/16 have mass of 74

• AM=70(9/16)+72(6/16)+74(1/16)=71amu

2.5 Periodic Table

Periods: the rows

Groups: the columns

• Group 1: alkali metals
• Group 2: alkaline earth metals
• Group 6: chalcogens
• Group 7: halogens
• Group 8: noble gases

2.6 Molecules, molecule compounds

Diatomic: molecules made of 2 atoms naturally (H, O, N, F, Cl, Br, I)

Molecular formula vs empirical formula: molecule is the actual # of atoms in a molecule while empirical is the smallest ratio

• Molecular: H2O2 Empirical: HO

Structural formula: drawing that shows how the atoms are joined but doesn’t show geometry

2.7 Ions, ionic compounds

ion: when an atom isn’t neutral, it either lost/gained electrons

• cation: positive ion, lost electrons
• anion: negative ion, gained electrons
• polyatomic ion: atoms joined as one molecule with an overall charge

ionic compound: cations and anions in an alternating crystal lattice. Always in empirical form

• formula unit: lowest whole # ratio of cations:anions

3.3 Formula weights

Formula weight: the sum of each atomic weight in a substance

• FW of H2O: 2(1.008)+1(16)=18.016 g

==%== ==composition: the mass contributed by each element==

• % comp= #atoms of element (atomic weight) /formula weight x100

Ex: calculate the % composition of C, H, and O for C12H22O11

1. C: 12(12.01)/342.296 x100=42.10%
2. H: 22(1.008)/342.296 x100=6.48%
3. O: 11(16)/342.296 x100=51.42%

Ex: calculate % composition of phosphorus in calcium phosphate Ca3(PO4)2

1. P: 2(30.97)/310.18 x100=19.96%

3.4: Avogadro’s number

Mole: 6.02x10^23

Calculate the # of H atoms in 0.350 mol of C6H12O6

=2.53x10^24 H atoms

Molar mass: moles of different elements have different masses

• 1 mole of Cl=35.45 amu but 1 mole of Au is 197amu

Calculate # moles of glucose in 5.380g sample

= 0.0298 moles of glucose

Calculate mass of 0.4333 moles of Ca(NO3)2

=71.06g Ca(NO3)2

5.23 g sample of glucose. Find # molecules

=1.75x10^22 molecules glucose

==3.5 Empirical formulas==

Empirical formula determination--Given percentages of each element

1. Base calculation on 100.g of compound. It’s easier
2. Determine # moles of each element for 100.g of compound
3. Divide each mole value by the smallest mole value to get the ratio
4. Multiply by the integer to get a whole number formula

Ex: adipic acid has 46.32% C, 43.84% O, 6.85% H

1. C: 46.32g →4.107 moles C O: 43.84g→2.74 moles O H: 6.85g →6.78 moles H
2. C: 4.107moles C/2.74 moles=1.50 O:2.74moles/2.74=1.00 H: 6.78moles H/2.74=2.50
3. Since not whole numbers, multiply everything by 2 →C3H5O2

Ex: Compound is 74% Hg and 26% Cl by mass. Find empirical

• 74g Hg and 26g Cl
• 74g Hg(1mole Hg/200.6g Hg)=0.369 moles Hg
• 26g Cl(1mole Cl/35.45g Cl)=0.733 moles Cl
• Hg:0.369moles Hg/0.369 =1 Hg
• Cl: 0.733moles Cl/0.369=2 Cl
• Empirical: HCl2

Ex: ascorbic acid is 40.92% C, 4.58% H, and 54.50% O. Find empirical formula

1. 40.92g C, 4.58g H, 54.50g O
2. 40.92g C(1mole C/12.01gC)=3.407 moles C
3. 4.58gH(1mole H/1.008g H)=4.54 moles H
4. 54.50g O(1 mole O/16g O)=3.406 moles O
5. C: 3.407g/3.406=1 C
6. H: 4.54 moles H/3.406=1.33 H
7. O: 3.406g O/3.406=1
8. Empirical: C3H4O3

finding the molecular formula from empirical formulas

==whole # multiple=molecular weight/empirical formula weight==

1. find formula mass of empirical formula
2. Find molar mass of the substance
3. substance molar mass/empirical molar mass
4. Multiply the empirical formula by that integer

Ex: empirical formula of adipic acid is C3H5O2 and the molar mass is 146g/mole. What is the molecular formula?

1. 146 g molar mass/73.07 g formula mass =2
2. 2(C3H5O2)= C6H10O4

Ex: C3H4O3 has an experimental mass of 76g.

1. 88.062g C3H4O3 /76g=2
2. 2(C3H4O3)=C6H8O6

Ex: C3H4 has an experimental mass of 121 amu, what’s molecular formula?

• 121amu/40.062amu=3
• 3(C3H4)=C9H12

Combustion analysis to find empirical formula

1. Use mass of CO2 to find the amount of C in organic substance
2. Use mass of H2O to find amount of H in organic substance
3. If there’s oxygen in the organic, subtract Cmass and Hmass to get Omass by itself
4. Once you have masses of each element, proceed like before, get mole substance ratios

Ex: menthol is composed of C, H, and O. A 0.1005g sample combusts, producing 0.2829g CO2 and 0.1159g H2O. What is the empirical formula if the molar mass of the organic is 156g/mole

1. 0.2829g CO2(mole CO2/44.01g CO2)(mole C/mole CO2)(12.01g C/mole C)=0.07715g C
2. 0.1159g H2O(mole H2O/18.016g H2O)(2 moles H/mole H2O)(1.008g H/mole H)=0.01288g H
3. 0.1005-0.0771-0.01288=0.01047g O
4. C: 0.07715g→0.006429 moles C
5. H: 0.01288g→0.01288 moles H
6. O: 0.01047g→0.0006544 moles O
7. C: 0.006429moles C/0.0006544=10 moles of C
8. H: 0.01288moles/0.0006544=20 moles of H
9. O: 0.0006544/0.0006544=1 mole of O
   10. Empirical formula C10 H20 O

Ex:Isopropyl alcohol has C, H, O. The combustion of 0.255g alcohol produces 0.561g CO2 and 0.306g H2O

1. 0.561g CO2(mole CO2/44g CO2)(1 mole C/mole CO2)(/12.01g C/mole C)=0.1531g C
2. 0.306g H2O(mole H2O/18.016g H2O)(2 mole H/mole H2O)(1.008g H/mole H)=0.0342g H
3. 0.255-0.1531-0.0342=0.0677 g O
4. 0.1531g C (mole C/12.01g C)=0.0127moles C
5. 0.0342g H(mole H/1.008g H)=0.0339 moles H
6. 0.0677g O(mole O/16g O)=0.00423 moles O
7. 0.0127moles C/0.00423=3 Carbon
8. 0.0339 moles H/0.00423=8 Hydrogen
9. Empirical=C3 H8 O

3.6 Quantitative Info from Balanced Equations

Coefficients: the relative # of molecules in a reaction

• Ex: 2H2 +O2 →2H2O shows 2 molecules of H2 reacting with 1 molecule of O2 to form 2 molecules H2O

Stoichiometrically equivalent quantities: used to convert between quantities

1. 1.57 mole O2 reactions w H2 to form H2O, how many moles H2 used?

• 1.57mole O2 x 2mole H2/mole O2 =3.14 mole H2

1. Determine how many g of H2O produced: C6H12O6 +6O2 →6CO2 +6H2O

• 1.00g glu x mole glu/180.156g glu x 6 mole H2O/mole glu x 18.016g H2O/mole H2O=0.600g H2O

3.7 Limiting Reactants

Limiting Reactant: the reactant that’s used up first. Once it’s gone, reaction can’t continue

1. N2 +3H2 →2NH3, 3moles N2, 6 moles H2

• 3 mole N2 x 2mole NH3/mole N2 =6mol NH3
• 6mole H2 x2mol NH3/3mole H2=4 mol NH3
• LR: H2, RIE: N2, N2 left over: 1 mole

Theoretical and Percent Yields

Theoretical yield: quantity of product calculated to form when all LR is consumed (100% completion, no error)

Experimental yield: amount of product actually obtained (will always be less due to error)

Percent yield: how much you got exp/theo x100%

Percent error: how far you were exp-theo/theo x100%

4.5 Concentrations of Solutions

Concentration: amt of solute dissolved in given quantity of solvent

• more solute=more concentrated

Molarity: concentration of moles/L

1. Calculate M of solution with 23.4g Na2SO4 to make 125ml

• 23.4g Na2SO4/0.125L =1.32M Na2SO4

1. What’s M of each ion in 0.025M Ca(NO3)2

• 0.025mol Ca(NO3)2/L x mole Ca2+/mole Ca(NO3)2=0.025mol Ca2+
• 0.025mol Ca(NO3)2/L x 2mole NO3-/mole Ca(NO3)2 =0.050 mol NO3-

1. 0.200M HNO3 solution. Calculate moles in 2.0 L

• 2.0L solution x 0.200mole HNO3/1L =0.40 moles HNO3

Dilution: adding water to make concentration lower

• C1V1 = C2V2

1. Want 250 ml of 0.100 M CuSO4 by diluting 1.00 M CuSO4

• C1= 1.00 M, V1=? C2=0.100M, V2=0.250L
• V1=(0.100M)(0.250L)/1.00M =.0250L or 25 ml of CuSO4 used

2. How many mL of 3.0M H2SO4 for 450mL 0.10M H2SO4

• C1=3.00M, V1=? C2=0.10M, V2=0.450L
• V1=(0.10M)(0.450L)/3.00M = 0.015 L or 15mL used

4.6 Solution Stoichiometry and Chem Analysis

If we know the solute M, we can used M and V to determine # moles

1. How many g of Ca(OH)2 to neutralize 25.0mL of 0.100 HNO3 Ca(OH)2 +2HNO3 →Ca(NO3)2 +2H2O

• 0.025 L x 0.100 mole NO3/L x mole Ca(OH)2/2mole HNO3 x 74.096g/mole Ca(OH)2 =0.092 g Ca(OH)2

Titrations: combining solution with unknown concentration w reagent of known concentration (standard sol)

Equivalence point: where stoich equivalent quantities are brought together

• Indicator: dye that changes color as passing equivalence point

1. 45.7mL of 0.5 M H2SO4 req. to neutralize 20.0mL of NaOH sol. What is M of NaOH?

• H2SO4 + 2NaOH →2H2O +Na2SO4
• 45.7mL H2SO4 x L/1000ml x 0.5 mole H2SO4/L =0.229 moles H2SO4
• 0.229 moles H2SO4 x 2mole NaOH/1mole H2SO4 =0.046 moles NaOH
• 0.046 moles NaOH/20ml x 1000ml/L =2.285 M NaOH