How to balance a chemical equation in Python 2.7 Using matrices
Question:
I have a college assignment where I must balance the following equation :
NaOH + H2S04 –> Na2S04 + H20
my knowledge of python and coding in general is extremely limited at the moment.
So far I have attempted to use matrices to solve the equation.
It looks like I am getting the solution a=b=x=y=0
I guess I need to set one of the variables to 1 and solve for the other three.
I’m not sure how to go about doing this,
I have had a search, it looks like
other people have used more sophisticated code and I’m not really able to follow it!
here’s what I have so far
#aNaOH + bH2S04 --> xNa2SO4 +y H20
#Na: a=2x
#O: a+4b=4x+y
#H: a+2h = 2y
#S: b = x
#a+0b -2x+0y = 0
#a+4b-4x-y=0
#a+2b+0x-2y=0
#0a +b-x+0y=0
A=array([[1,0,-2,0],
[1,4,-4,-1],
[1,2,0,-2],
[0,1,-1,0]])
b=array([0,0,0,0])
c =linalg.solve(A,b)
print c
0.0.0.0
Answers:
The problem is that you have constructed a linear system with b being a zero-vector. Now for such system there is always the straight-forward answer that all variables are zeros as well. Since multiplying a number with zero and adding zeros up results always in zeros.
A solution might be to assign 1 to a variable. Take for instance a. If we assign a = 1, then we will get b, x and y in function of a being 1.
So now or linear system is:
B X Y | #
2 |1 # A = 2X
-4 4 1 |1 # A+4B = 4X+4Y
-2 2 |1 # A+2B = 2Y
-1 1 0 |0 # B = X
Or putting it into code:
>>> A = array([[0,2,0],[-4,4,1],[-2,0,2],[-1,1,0]])
>>> B = array([1,1,1,0])
>>> linalg.lstsq(A,B)
(array([ 0.5, 0.5, 1. ]), 6.9333477997940491e-33, 3, array([ 6.32979642, 2.5028631 , 0.81814033]))
So that means that:
A = 1, B = 0.5, X = 0.5, Y = 1.
If we multiply this by 2, we get:
2 NaOH + H2S04 -> Na2S04 + 2 H20
Which is correct.
I referred to Solve system of linear integer equations in Python which translated into
# Find minimum integer coefficients for a chemical reaction like
# A * NaOH + B * H2SO4 -> C * Na2SO4 + D * H20
import sympy
import re
# match a single element and optional count, like Na2
ELEMENT_CLAUSE = re.compile("([A-Z][a-z]?)([0-9]*)")
def parse_compound(compound):
"""
Given a chemical compound like Na2SO4,
return a dict of element counts like {"Na":2, "S":1, "O":4}
"""
assert "(" not in compound, "This parser doesn't grok subclauses"
return {el: (int(num) if num else 1) for el, num in ELEMENT_CLAUSE.findall(compound)}
def main():
print("nPlease enter left-hand list of compounds, separated by spaces:")
lhs_strings = input().split()
lhs_compounds = [parse_compound(compound) for compound in lhs_strings]
print("nPlease enter right-hand list of compounds, separated by spaces:")
rhs_strings = input().split()
rhs_compounds = [parse_compound(compound) for compound in rhs_strings]
# Get canonical list of elements
els = sorted(set().union(*lhs_compounds, *rhs_compounds))
els_index = dict(zip(els, range(len(els))))
# Build matrix to solve
w = len(lhs_compounds) + len(rhs_compounds)
h = len(els)
A = [[0] * w for _ in range(h)]
# load with element coefficients
for col, compound in enumerate(lhs_compounds):
for el, num in compound.items():
row = els_index[el]
A[row][col] = num
for col, compound in enumerate(rhs_compounds, len(lhs_compounds)):
for el, num in compound.items():
row = els_index[el]
A[row][col] = -num # invert coefficients for RHS
# Solve using Sympy for absolute-precision math
A = sympy.Matrix(A)
# find first basis vector == primary solution
coeffs = A.nullspace()[0]
# find least common denominator, multiply through to convert to integer solution
coeffs *= sympy.lcm([term.q for term in coeffs])
# Display result
lhs = " + ".join(["{} {}".format(coeffs[i], s) for i, s in enumerate(lhs_strings)])
rhs = " + ".join(["{} {}".format(coeffs[i], s) for i, s in enumerate(rhs_strings, len(lhs_strings))])
print("nBalanced solution:")
print("{} -> {}".format(lhs, rhs))
if __name__ == "__main__":
main()
which runs like
Please enter left-hand list of compounds, separated by spaces:
NaOH H2SO4
Please enter right-hand list of compounds, separated by spaces:
Na2SO4 H2O
Balanced solution:
2 NaOH + 1 H2SO4 -> 1 Na2SO4 + 2 H2O
Very well done. However, when I tested this snippet on the following equation taken from David Lay’s Linear Algebra textbook, the 5th edition I received a sub-optimal solution that can be further simplified.
On p. 55, 1.6 Exercises check ex 7.:
NaHCO_3 + H_3C_6H_5O_7 –> Na_3C_6H_5O_7 + H_2O + CO_2
Your snippet returns:
Balanced solution:
15NaHCO3 + 6H3C6H5O7 -> 5Na3C6H5O7 + 10H2O + 21CO2
The correct answer is:
3NaHCO_3 + H_3C_6H_5O_7 -> Na_3C_6H_5O_7 + 3H_2O + 3CO_2
You can use this solution. It works with any chemical equation. The last coefficient can be calculated with a row where b[i]!=0
H2SO4+NaOH−−>Na2SO4+H2OH2SO4+NaOH−−>Na2SO4+H2O
a=np.array([[2,1,0],[1,0,-1],[4,1,-4],[0,1,-2]])
b=np.array([2,0,1,0])
x=np.linalg.lstsq(a,b,rcond=None)[0]
print(x)
y=sum(x*a[0])/b[0]
print("y=%f"%y)
out:
[0.5 1. 0.5]
y=1.000000
I have a college assignment where I must balance the following equation :
NaOH + H2S04 –> Na2S04 + H20
my knowledge of python and coding in general is extremely limited at the moment.
So far I have attempted to use matrices to solve the equation.
It looks like I am getting the solution a=b=x=y=0
I guess I need to set one of the variables to 1 and solve for the other three.
I’m not sure how to go about doing this,
I have had a search, it looks like
other people have used more sophisticated code and I’m not really able to follow it!
here’s what I have so far
#aNaOH + bH2S04 --> xNa2SO4 +y H20
#Na: a=2x
#O: a+4b=4x+y
#H: a+2h = 2y
#S: b = x
#a+0b -2x+0y = 0
#a+4b-4x-y=0
#a+2b+0x-2y=0
#0a +b-x+0y=0
A=array([[1,0,-2,0],
[1,4,-4,-1],
[1,2,0,-2],
[0,1,-1,0]])
b=array([0,0,0,0])
c =linalg.solve(A,b)
print c
0.0.0.0
The problem is that you have constructed a linear system with b being a zero-vector. Now for such system there is always the straight-forward answer that all variables are zeros as well. Since multiplying a number with zero and adding zeros up results always in zeros.
A solution might be to assign 1 to a variable. Take for instance a. If we assign a = 1, then we will get b, x and y in function of a being 1.
So now or linear system is:
B X Y | #
2 |1 # A = 2X
-4 4 1 |1 # A+4B = 4X+4Y
-2 2 |1 # A+2B = 2Y
-1 1 0 |0 # B = X
Or putting it into code:
>>> A = array([[0,2,0],[-4,4,1],[-2,0,2],[-1,1,0]])
>>> B = array([1,1,1,0])
>>> linalg.lstsq(A,B)
(array([ 0.5, 0.5, 1. ]), 6.9333477997940491e-33, 3, array([ 6.32979642, 2.5028631 , 0.81814033]))
So that means that:
A = 1, B = 0.5, X = 0.5, Y = 1.
If we multiply this by 2, we get:
2 NaOH + H2S04 -> Na2S04 + 2 H20
Which is correct.
I referred to Solve system of linear integer equations in Python which translated into
# Find minimum integer coefficients for a chemical reaction like
# A * NaOH + B * H2SO4 -> C * Na2SO4 + D * H20
import sympy
import re
# match a single element and optional count, like Na2
ELEMENT_CLAUSE = re.compile("([A-Z][a-z]?)([0-9]*)")
def parse_compound(compound):
"""
Given a chemical compound like Na2SO4,
return a dict of element counts like {"Na":2, "S":1, "O":4}
"""
assert "(" not in compound, "This parser doesn't grok subclauses"
return {el: (int(num) if num else 1) for el, num in ELEMENT_CLAUSE.findall(compound)}
def main():
print("nPlease enter left-hand list of compounds, separated by spaces:")
lhs_strings = input().split()
lhs_compounds = [parse_compound(compound) for compound in lhs_strings]
print("nPlease enter right-hand list of compounds, separated by spaces:")
rhs_strings = input().split()
rhs_compounds = [parse_compound(compound) for compound in rhs_strings]
# Get canonical list of elements
els = sorted(set().union(*lhs_compounds, *rhs_compounds))
els_index = dict(zip(els, range(len(els))))
# Build matrix to solve
w = len(lhs_compounds) + len(rhs_compounds)
h = len(els)
A = [[0] * w for _ in range(h)]
# load with element coefficients
for col, compound in enumerate(lhs_compounds):
for el, num in compound.items():
row = els_index[el]
A[row][col] = num
for col, compound in enumerate(rhs_compounds, len(lhs_compounds)):
for el, num in compound.items():
row = els_index[el]
A[row][col] = -num # invert coefficients for RHS
# Solve using Sympy for absolute-precision math
A = sympy.Matrix(A)
# find first basis vector == primary solution
coeffs = A.nullspace()[0]
# find least common denominator, multiply through to convert to integer solution
coeffs *= sympy.lcm([term.q for term in coeffs])
# Display result
lhs = " + ".join(["{} {}".format(coeffs[i], s) for i, s in enumerate(lhs_strings)])
rhs = " + ".join(["{} {}".format(coeffs[i], s) for i, s in enumerate(rhs_strings, len(lhs_strings))])
print("nBalanced solution:")
print("{} -> {}".format(lhs, rhs))
if __name__ == "__main__":
main()
which runs like
Please enter left-hand list of compounds, separated by spaces:
NaOH H2SO4
Please enter right-hand list of compounds, separated by spaces:
Na2SO4 H2O
Balanced solution:
2 NaOH + 1 H2SO4 -> 1 Na2SO4 + 2 H2O
Very well done. However, when I tested this snippet on the following equation taken from David Lay’s Linear Algebra textbook, the 5th edition I received a sub-optimal solution that can be further simplified.
On p. 55, 1.6 Exercises check ex 7.:
NaHCO_3 + H_3C_6H_5O_7 –> Na_3C_6H_5O_7 + H_2O + CO_2
Your snippet returns:
Balanced solution:
15NaHCO3 + 6H3C6H5O7 -> 5Na3C6H5O7 + 10H2O + 21CO2
The correct answer is:
3NaHCO_3 + H_3C_6H_5O_7 -> Na_3C_6H_5O_7 + 3H_2O + 3CO_2
You can use this solution. It works with any chemical equation. The last coefficient can be calculated with a row where b[i]!=0
H2SO4+NaOH−−>Na2SO4+H2OH2SO4+NaOH−−>Na2SO4+H2O
a=np.array([[2,1,0],[1,0,-1],[4,1,-4],[0,1,-2]])
b=np.array([2,0,1,0])
x=np.linalg.lstsq(a,b,rcond=None)[0]
print(x)
y=sum(x*a[0])/b[0]
print("y=%f"%y)
out:
[0.5 1. 0.5]
y=1.000000