如果我在Python中有Molar质量，我如何找到一种化合物的克呢？

Question

我有一些困难，试图从莫拉尔质量转换回克。该程序以x克(用户输入)的速度输出化合物A的摩尔质量。如果我有10克SiO2，那么Si和O2是多少克？

我将如何编码用户的输入，比方说10g SiO2，并将其转换为需要多少Si和g O？

10 gSi 2/M SiO2 * MSi =gSi

2*(10克SiO2 /M SiO2) * MO = gO (乘以2，因为O2)

M= Molar质量

G=克

M=摩尔

import Periodic_Table_new as pt

print("Enter info for Compound A\n")
compoundA = pt.getInput()
gramTotal = float(input("Please enter total grams of Compound A: "))
moles = pt.convertToMoles(gramTotal,compoundA)
moleMass = moles/compoundA
print('\n')
print("The Molecular Mass of compound A at " + str(gramTotal) + "grams is : " + str(moleMass) + "\n\n")

Periodic_Table_new.py

def getInput():
    compoundName = "" 


ix = True
total = 0
while ix:

    elementName = str(input("Please Enter Element: "))

    amountNum = int(input("Please Enter Amount of Element: "))

    elementWeight = float(input("Please enter Element Weight: "))
    elementWeight = amountNum * elementWeight

    total = total + elementWeight

    compoundName = compoundName + elementName + str(amountNum)
    print(compoundName)

    userInput = input("Would you like to repeate? \n Y/N? ")

    if userInput == 'N' or userInput == 'n':
        ix = False
print(compoundName)
print(total)
return total

def convertToMoles(gOc,c):
moles = gOc/c
return moles

def molesToGrams():
p = moles * n * m

Answer 1

基于我对你所期望的结果的理解，我想出了一些不同的方法。这个版本将使用原子质量字典、解析复合字符串的regex和进行数学运算的循环。它相当灵活，但它需要你建立原子质量字典，这应该不难，只是有点乏味。它对评论中提到的输入有一些要求。

import re

atomic_masses = {'Si': 28.0855, 'O': 15.999}

compound = input('Please enter a compound and amount: ')

amount_pat = re.compile(r'(\d+)g')  # compound mass must end in the letter g
element_pat = re.compile(r'([A-Z][a-z]?\d*)')  # elemental symbols must be properly capitalized (Si, not si)
sym_pat = re.compile(r'[A-Z][a-z]?')  # elemental symbols must be properly capitalized (Si, not si)
qty_pat = re.compile(r'\d+')

mass = int(amount_pat.search(compound)[1])

elements = []
# finds each element in the compound and makes a list of (element, parts) tuples
for element in element_pat.finditer(compound):
    element = element[0]
    symbol = sym_pat.search(element)[0]
    if any(c.isdigit() for c in element):
        qty = int(qty_pat.search(element)[0])
    else:
        qty = 1

    elements.append((symbol, qty))

# Calculates and prints total Molecular Mass for the compund
molecular_mass = sum(el[1] * atomic_masses[el[0]] for el in elements)
print(f'\nTotal Molecular Mass: {molecular_mass}\n')

# Calculates and prints the mass for each element
for tup in elements:
    unit_mass = (mass / molecular_mass) * atomic_masses[tup[0]] * tup[1]
    print(f'{tup[0]}: {unit_mass:.4f}g')

示例输出：

Please enter a compound and amount: 10g SiO2

Total Molecular Mass: 60.0835

Si: 4.6744g
O: 5.3256g

编辑：

为了适应输入的十进制质量，我们可以将amount_pat内部的正则表达式更改为r'(\d*.\d+)g'。现在它将接受诸如.003g或0.003g之类的值。它仍然不需要kg或mg，但是如果您查找内置的re包，您可以了解regex是如何工作的，并从那里更改它。regex101也是regex的一个很好的资源，因为它允许尝试和错误的学习方法。我还在int赋值语句中将mass更改为float

然而，我确实改变了元素质量循环来调整镁和千克。如果您希望进一步扩展它，只需采用我在这里开始的模式，并进一步扩展它：

for tup in elements:
    unit_mass = (mass / molecular_mass) * atomic_masses[tup[0]] * tup[1]
    if unit_mass > 1000:
        unit_mass /= 1000
        unit = 'kg'

    elif unit_mass < .01:
        unit_mass *= 1000
        unit = 'mg'

    else:
        unit = 'g'

    print(f'{tup[0]}: {unit_mass:.4f}{unit}')

新示例输出：

Please enter a compound and amount: .003g H2O

Total Molecular Mass: 18.01488

H: 0.3357mg
O: 2.6643mg

Answer 2

你将需要元素的结尾，这是缺少的部分。

如果没有输入魔法，下面是10g SiO2的一个示例：

target=10
elem=[['Si',1,28.09],['O',2,16]]

total=0
for e in elem:
  total=total+e[1]*e[2]
print('molar mass of compound:',total)

moles=target/total
print('moles:',moles)

for e in elem:
  print(e[0],moles*e[1],'mol',moles*e[1]*e[2],'g')

Answer 3

计算复合中元素的摩尔质量和质量百分比

我使用比伐仑包来实现您想要的输出。这是我的密码

from pyvalem.formula import Formula
import numpy as np

def give_compound_details(compound, given_gram):

    # Dictionary of element of mole of a compound
    ele_mol_dict = Formula(compound).atom_stoich

    # Create a list comprehensioon to get the mass per ele
    mass_per_ele = [Formula(ele).mass * ele_mol_dict[ele] for ele in ele_mol_dict.keys()]

    # Create dic for ele and mass
    ele_mass_dict = dict(zip(list(Formula(compound).atom_stoich),mass_per_ele))

    # Create a list comprehension and dictionary for mass percentage per element of a compound
    ele_percent_mass_dict = dict(zip(list(ele_mol_dict.keys()),[(ele_mass_dict[ele] / Formula(compound).mass * 100) for ele in ele_mass_dict.keys()]))

    
    print(f"Total Molecular mass of {compound}: {Formula(compound).mass} g\n")  
    for key, value in ele_mass_dict.items():
        print(f"Molar Mass of {key}: {value} g")
    print("")
    for key, value in ele_percent_mass_dict.items():
        print(f"Mass Percentage of {key}: {round(value,2)} %")

    print("")
    print(f"If molecular mass of {compound} is {given_gram} g, therefore")

    for key, value in ele_percent_mass_dict.items():
        print(f"Mass of {key}: {round((value/100)*given_gram,4)} g")

# MAIN PROGRAM
is_running = True
while is_running:
    print("Computing the Molar Mass and Mass Percentage of Elements within a Compound")
    print("")
    compound = input("Compound: ")
    given_gram = float(input(f"How much {compound} do you have in grams? "))
    print("\nRESULT\n")
    function = give_compound_details(compound, given_gram)
    
    print("---------------------------------------")
    ask = input('Do you want to try again? \nType "y" is YES, type "n" if NO')
    if ask.islower() == 'n':
        is_running = False
    print("---------------------------------------")

下面是示例输出：输出截图