Streptomyces:Other Bits/Useful Molecular and Chemical Equations
Other Bits  Useful Molecular and Chemical Equations
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Useful Molecular and Chemical EquationsFormula Weight & Molecular Weight Formula weight (FW) and molecular weight (MW) are calculated by summing the atomic weights (AW measured in atomic mass units, amu) of the individual atoms.
Avogadro’s Number & mole 1 mole of atoms / molecules has a mass equal to the atomic / molecular weight in grams. e.g. 1 mole (1mol) NaCl is the number of molecules in 58.44g of NaCl. (1mol NaCl = 58.44g) Avogadro’s number is the number of atoms / molecules in 1 mole of any substance, which is equal to 6.02214x10^{23}.
Molarity – Molar Concentration Molarity is the number of moles of solute per litre of solution. e.g. 6 molar (6M) HCl is equal to 6 moles (6mol) of HCl per litre (L). (6M HCl = 6mol/L)
Based on the previous two equations:
Primer Calculations Primers are dissolved in sterile distilled water (sdH_{2}O) to a concentration of 500pmol. Use one of the following to determine what volume of sdH_{2}O to use:
Weight / mole Percentage The percentage weight of an element in a compound is calculated using the atomic weight and formula weight.
[math]\displaystyle{ \frac {AW}{FW}*100=%weight }[/math] Percentage weight of Cl in HCl: [math]\displaystyle{ Cl=\frac {35.45}{36.45}=97.26% }[/math]
Similarly, mole percentage is a ratio.
[math]\displaystyle{ \frac {x}{T}*100=%mole }[/math] Percentage mole of Cl in HCl: [math]\displaystyle{ Cl=\frac {1}{2}*100=50% }[/math] Density and Specific Gravity Density is the mass of a substance per volume.
Specific gravity is a unitless ratio, so for all purposes; SG ≡ D. Cubic centimetres are equivalent to millilitres; cc ≡ mL.
D_{2} = Density of H_{2}O @ 4°C = 1.00g/cc Molarity, Specific Gravity and Percentage Composition Calculating Molarity from specific gravity and percentage composition:
Percentage composition means xg of pure compound per 100g of solution, i.e. 37g/100ml = 37%. To calculate the molarity, the mass of pure compound is needed; however the solution’s specific gravity needs to be taken into account, and the volume; which we’ll take to be 1L.
Where 1L = 1000cc For HCl: [math]\displaystyle{ m=\frac {1.18*1000*37}{100}=436.6g }[/math] Therefore: [math]\displaystyle{ M=\frac {436.6}{1*36.45}=11.97mol/L }[/math]
