SI Units Unit Conversion • Problem : The density of oxygen at room • Problem : The SI unit for density is temperature is about 1.3 kg/m 3 . Express its – g/m 3 density in g/cm 3. 3 – 0.1Kg/m – 0.013 – m/Kg 3 – 13 SI base units – g/mL – 1300 – Kg/m 3 – 130 – 0.0013 • Answer : Kg/m 3 • Solution : 1 kg = 10 3 g. 1 m 3 = 100x100x100cm 3 = 10 6 cm 3 • Answer : 1.3 kg/m 3 = 1.3 x 10 3 / 10 6 = 0.0013 g/cm 3

Energy Unit Conversion Kelvin vs Celsius • Problem : 1 Joule is close to the following value: • Problem : Patient’s body temperature is – 4.2 kcal determined to be 313 K. The patient is most likely – 2.092 cal – healthy – 0.24 cal – sick – 8.314 cal – dead – 1.035 cal – 4.184 cal • Answer : 313 K ~ 313-273 = 40 C; the patient has • Answer : 1 Joule = 0.24 cal; 1 cal = 4.184 Joule a fever therefore he/she is sick.

Gas Constant Energy Scale • Problem : Calculate RT at 0 °C in cal/mol • Problem : What is an approximate value of the energy of a covalent bond? • Answer : Many faces of the gas constant R: – 50 kcal/mol to 100 kcal/mol R ≈ 8.314 J / (K ⋅ mol) – exactly 10.45 kcal/mol ≈ 19 ⋅ R 5.189×10 eV / (K mol) – 1 to 2 kcal/mol R ≈ 0.082 L⋅atm / (K ⋅ mol) R ≈ 1.986 cal / (K ⋅ mol) – about 0.1 kcal/mol and T = 0 °C ≈ 273 K; therefore RT = 1.986 x 273 – 1 to 5 kcal/mol = 542.178 cal/mol ~ 0.54 kcal/mol • Answer : 50 kcal/mol to 100 kcal/mol

• At room temp , RT ~ 0.6 kcal/mol Biological scale Biological Scale cntd. • Problem: The cell membrane and the membranes surrounding inner cell organelles are phospholipid bilayers • Problem : The distance between centers of two about ____ thick covalently bonded carbon atoms is close to: – 5 nm – 1.5 nm – 100 pm (pico meter) – 1 A – 1.5 µm – 5 A – 1.5 A – 50 nm – 0.015 nm – 5 µm – 150 A • Answer: 5 nm. • Hint: membrane is a bilayers of phospholipids. Each lipid has • Answer : 1.5 A, same as 0.15 nm or 150 pm a head (~ 5 covalent bonds tall) and a hydrocarbon chain (~ 17-20 covalent bonds long). 2 x 25 x 1.5 A (for the length of ~1.7A C-C bond) = 75 A = 7.5 nm. The actual answer is about the same order, but smaller, because bonds are connected at 3 A ~120 o angles. ~1.5A 4 A 5 A

Molar Amount vs Weight Concentrations, volumes, molar amounts… • Problem: Estimate the weight of a 0.5 mmol sample of Fomepizole. • Concentration = molar amount / volume – 0.04 g – measured in mol/L ≡ M, also mM, µM, nM, pM etc. – 0.004 mg – molar amount = volume x concentration – 0.5 µg – volume = molar amount / concentration – 82 g • MW = Mass / molar amount – 4.05 g – measured in g/mol ≡ Da • Answer: 0.04 g – Mass = MW x molar amount • Solution: MW(Fomepizole) is ~ – molar amount = mass / MW 4x12(C)+2x14(N)+5(H) ~ 81. 0.5 mmol x 81 g/mol = 0.5 x 10 -3 x 81 g/mol ~ 0.04 g

Molar Amount vs Weight cntd. Concentration vs Amount

• Problem : Albumin is the most abundant serum • Problem : What is the total molar amount of a protein. Its concentration in plasma ranges from 30 compound in 0.3 mL of 1 mM solution of that to 50 g/L. Given that MW for albumin is 67 kDa, compound? estimate its molarity. – 0.3 µmol – 440-740 nM – impossible to tell because the MW of the compound is not – 4-7 uM given – 440-740 uM – 3.33 mol – 4-7 mM – 0.3 mg – 44-74 mM – 0.3 mol • Solution : Molar amount = concentration x volume -4 • Answer : 33.5 g/L / 67000 = 5 x 10 mol/L = 0.5 mM = (1 x 10 -3) x (0.3 x 10 -3) = 0.3 x 10 -6 = 0.3 µ mol or 500 uM; the correct range includes this number. 23 Avogadro Number ~ 6x10 Degrees of Freedom • Problem: The approximate mass of one • DF are the variables capable of storing kinetic or molecule of Penciclovir is 4.2 x 10 -22 g. potential energy Calculate the molecular weight of the drug. • # of DF discretely increases with T – 252 g/mol • DF of a molecule in gas phase:

– 172 g/mol – 3 trans – 326 g/mol – + 0, 2, or 3 rot – + N (excited torsional variables only) – 472 g/mol tor – + 2N (excited vibrations only; >vibrational T) – 504 g/mol vib – Most torsional variables ARE excited at 300 K • Solution : – Most bond vibrations are NOT excited at 300 K -22 23 MW = 4.2 x 10 x 6 x 10 ~ 252 g/mol (Exception: I 2)

Degrees of freedom below vib. T. Degrees of freedom 3 5 • Problem: Determine the number of degrees of (translational only) (3 trans + 2 rot) freedom of adenosine triphosphate (ATP) in gas phase below vibrational temperature. – 6 – 4 – 2 6 Many – 3 (3 trans + 3 rot) (3 trans + 3 rot + conf) – 1 – 5 – the correct answer not given – infinitely many • Answer: The correct answer is not given; it is finite but significantly greater than 6

Volume of 1 mole of gas Kinetic Energy & Gas Law • Problem : Estimate the volume of 1 mole of ° ° • E = ½ mv 2 = 3/2 RT nitrous oxide at 0 C and at 27 C – Root mean square velocity: v = Sqrt (3RT/m) • Solution : PV = nRT • Gas Law: PV = nRT – V = RT/P (because n = 1) – V = nRT/P – Many faces of R: R = 0.082 L⋅atm / (K⋅mol) – At 273 K, RT ≈ 22.4 L⋅atm / mol; V ≈≈≈ 22.4 L – At 300 K, RT ≈ 24.6 L⋅atm / mol; V ≈≈≈ 24.6 L – Applies to any gas Comparative sizes of pharmacologically Energy vs Temperature relevant molecules • Problem : The Celsius temperature in a storage room was increased from 25C to 50C. How much did the average kinetic energy of molecules in the room change? Mark the closest answer. – increased by 2½ times – decreased – increased by 8 % – increased by 2 times – The increase cannot be calculated without knowing the molecular mass • Answer : The kinetic energy is proportional to Kelvin temperature, which increased from 273 + 25 = 298 to 273 + 50 = 323. 323/298 ~ 1.08, i.e. the temperature, as well as the average kinetic energy, increased by 8%. • RMS velocity will increase by ~ 4% (!)