Variables and Arithmetic Operators

Variables and Arithmetic Operators#

Variables for saving values#

To save a value, we assign them to a variable for later use.

The syntax for assigning variables is: variable_name = variable_value.

Use print(variable_name) to print the specified variable.

Tips:

  • Choose informative names for variables.

  • Use comment lines (lines with #) to express the units of the variable or to describe the meaning of the variable.

Arithmetic operators for calculations#

Common arithmetic operators are:

  • + for addition

  • - for subtraction

  • * for multiplication

  • \ for division

  • ** for power operations.

Examples#

Please pay attention to the use of comments (with #) to express the units of variables or to describe the meaning of commands.

Example

How many grams of solid NaOH (40.0 g/mol) are required to prepare 500 ml of a 40 mM solution?

Use \(Mass\) \((g)\) = \(Molar\) \(Concentration\) \((mol/l)\) x \(Volume\) \((l)\) \(Molecular\) \(Weight\) \((g/mol)\)

V = 0.5   #l
M = 0.04   #mol/l
MW = 40.0   #g/mol

m = (V * M) * MW   #(l * mol/l) * g/mol = g

print(m)   #print the value that we calculated

0.8

Example

How many ml of a 1 M Tris solution do you need to make 5 ml of a 0.25 M solution?

Use \(C_{initial}\) \(V_{initial}\) = \(C_{final}\) \(V_{final}\)

with

  • \(V_{initial}\) = volume of stock solution needed to make the new solution

  • \(C_{initial}\) = concentration of the stock solution

  • \(V_{final}\) = final volume of the new solution

  • \(C_{final}\) = final concentration of the new solution

Vfinal = 5   #ml
Cfinal = 0.25   #mol/l
Cinitial = 1   #mol/l

Vinitial = (Cfinal * Vfinal) / Cinitial   #(mol/l * ml) / mol/l = ml

print(Vinitial)   #print the value that we calculated

1.25

Exercises#

Exercise

You need 250 ml 50 mM NaCl. You have a 2 M NaCl stock solution. How many ml of 2 M NaCl do you need?

Solution

Here’s one possible solution.

Vfinal = 0.25   #l
Cfinal = 0.05   #mol/l
Cinitial = 2   #mol/l

Vinitial = (Cfinal * Vfinal) / Cinitial   #(l * mol/l) / mol/l = l
Vinitialml = Vinitial*1000   #l/1000 = ml

print(Vinitialml)   #print the value that we calculated

Exercise

You have 1 g of ATP (551.1 g/mol). You need a 100 mM solution. How much solution (in ml) can you make?

Solution

Here’s one possible solution.

m = 1   #g
M = 0.1   #mol/l
MW = 551.1   #g/mol

V = (m / MW) / M   #(g * g/mol) * mol/l = l
Vml = V*1000   #l/1000 = ml

print(Vml)   #print the value that we calculated
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