Question

我在尝试将摩尔质量转换回克时遇到了一些困难。该程序以x克（用户输入）吐出化合物A的摩尔质量。如果我有10克的SiO2，那么Si和O2的克数（相应地）是多少？

我如何编写用户输入的10g二氧化硅并将其转换为需要多少的Si和g的O？

10gSiO 2 / M SiO 2 * MSi = gSi

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

M =摩尔质量

g =克

m = mole

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

你需要最后的元素，那就是缺少的部分。

没有输入法术，这里是10g SiO2的例子：

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 2

根据我对你理想结果的理解，我提出了一些不同的方法。这个版本将使用原子质量字典，正则表达式解析复合字符串，以及循环来进行数学运算。它非常灵活，但它需要你建立原子质量的字典，这应该不难，只是有点单调乏味。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内的正则表达式更改为r'(\d*.\d+)g'。现在它将接受.003g或0.003g等值。它仍然不需要kg或mg，但是如果你查找内置的re包，你就可以了解regex如何工作并从那里改变它。 regex101也是正则表达式的一个很好的资源，因为它允许使用试错法进行学习。我还在int转让声明

中将float更改为mass 然而，我确实改变了元素质量循环以调整mg和kg。如果你想进一步扩展它，只需采取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

如果我在Python中使用摩尔质量，如何找到化合物的克数？

2 个答案:

编辑：