Stoichiometry
IGCSE Chemistry Topic 3 15:33 English narration · English + 中文 subtitles burned in
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Here is a chemist's problem.
化学家有一个难题。
A single atom is far too small to see, and there are billions upon billions in the tiniest speck.
单个原子太小了，根本看不见，而最微小的一点点里就有亿万个。
So how do you count them?
那你怎么数它们呢？
You can't — so instead, you weigh them.
你数不了——所以，我们改为称它们的重量。
Chemists use a clever unit called the mole.
化学家使用一个巧妙的单位，叫做摩尔。
One mole is a fixed, enormous number of particles, always the same.
一摩尔是固定的、庞大的粒子数目，永远相同。
It lets us count atoms simply by weighing — and it turns chemistry into arithmetic.
它让我们只靠称重就能数出原子——把化学变成了算术。
Welcome to stoichiometry — the chemistry of amounts.
欢迎来到化学计量学——研究物质用量的化学。
Formulae, balanced equations, and the mole.
化学式、配平方程式，还有摩尔。
Let's begin.
让我们开始吧。
First, formulae.
首先，化学式。
A formula tells you which atoms are in a substance, and how many.
化学式告诉你一种物质里有哪些原子，各有多少个。
Glucose is C-six H-twelve O-six: six carbons, twelve hydrogens, and six oxygens.
葡萄糖是碳六氢十二氧六：六个碳、十二个氢、六个氧。
For an ionic compound, you can build the formula from the ion charges.
对于离子化合物， 你可以根据离子电荷来写化学式。
The positive and negative charges must cancel.
正负电荷必须相互抵消。
Take aluminium, charge three plus, and oxygen, charge two minus.
以铝为例，电荷是正三； 氧，电荷是负二。
The charges cross over to become the amounts — two aluminium and three oxygen.
电荷交叉过来，变成原子的个数——两个铝、三个氧。
Aluminium oxide, Al-two O-three.
氧化铝，铝二氧三。
You must know two kinds of formula.
你必须知道两种化学式。
The molecular formula is the actual number of each atom in one molecule.
分子式是一个分子里每种原子的实际个数。
For glucose that is C-six H-twelve O-six.
葡萄糖是碳六氢十二氧六。
The empirical formula is the simplest whole-number ratio of the atoms.
实验式是原子的最简整数比。
Divide those counts by six, and glucose becomes C H-two O.
把那些个数都除以六，葡萄糖就变成碳氢二氧。
Same compound, two different formulae — one tells you the real molecule, the other only the ratio.
同一种化合物，两种不同的化学式——一个告诉你真实分子，另一个只告诉比例。
Look at the charges crossing over.
看电荷怎样交叉。
Aluminium is three plus, oxygen is two minus.
铝是正三，氧是负二。
The three swaps onto oxygen, the two swaps onto aluminium — so you write Al-two O-three.
三换到氧上，二换到铝上——所以写成铝二氧三。
The compound has no overall charge: two times three plus, and three times two minus, cancel.
化合物整体不带电：两个正三，加上三个负二，正好抵消。
Try sodium and oxide: one sodium is one plus, oxide is two minus, so you need two sodium ions.
再试钠和氧：钠是正一， 氧是负二，所以需要两个钠离子。
The formula is Na-two O.
化学式是钠二氧。
Memorise common ions: sodium one plus, magnesium two plus, aluminium three plus; chloride one minus, oxide two minus, sulfate two minus.
记住常见离子：钠正一、镁正二、 铝正三；氯负一、氧负二、硫酸根负二。
A chemical equation shows reactants turning into products.
化学方程式表示反应物变成生成物。
In words: magnesium plus oxygen makes magnesium oxide.
用文字：镁加氧生成氧化镁。
In symbols, it must be balanced — the same number of each atom on both sides, because atoms are never created or destroyed.
用符号，它必须配平—— 两边每种原子的数目相同，因为原子既不会被创造，也不会被消灭。
Two magnesium, plus one oxygen molecule, makes two magnesium oxide.
两个镁，加一个氧分子， 生成两个氧化镁。
Count them: two magnesiums and two oxygens on each side.
数一数：每边两个镁、两个氧。
The golden rule: to balance, you may only change the big numbers in front — never the small numbers inside a formula.
黄金法则：配平时， 你只能改前面的大数字——绝不能改化学式里面的小数字。
A balanced equation can also show the physical state of each substance.
配平的方程式还可以标出每种物质的物理状态。
You add a letter in brackets after the formula.
在化学式后面的括号里加一个字母。
Solid is s, liquid is l, gas is g, and aqueous — dissolved in water — is a-q.
固体是固，液体是液，气体是气，水溶液——溶解在水里——是水溶。
Zinc solid plus hydrochloric acid aqueous makes zinc chloride aqueous and hydrogen gas.
锌固体加盐酸水溶液，生成氯化锌水溶液和氢气。
Those little letters are state symbols.
这些小字母就是状态符号。
Examiners expect them when the question asks for a full equation.
题目要求完整方程式时，阅卷老师会期望看到它们。
Sometimes you write an ionic equation.
有时你会写离子方程式。
It shows only the ions that actually change.
它只显示真正发生变化的离子。
Ions that stay the same on both sides are spectator ions — leave them out.
两边保持不变的离子是旁观离子—— 把它们删掉。
When an acid reacts with an alkali, the useful change is hydrogen ion aqueous plus hydroxide ion aqueous making water liquid.
当酸与碱反应时，有用的变化是氢离子水溶液加氢氧根离子水溶液生成液态水。
Sodium and chloride may be present, but they are spectators.
钠和氯也许在场，但它们是旁观者。
The short ionic equation is the one that earns the marks.
短的离子方程式才是拿分的那一个。
Every atom has a relative atomic mass — its average mass compared to a carbon atom.
每个原子都有一个相对原子质量——它与一个碳原子相比的平均质量。
Hydrogen is one, carbon is twelve, oxygen is sixteen.
氢是一，碳是十二，氧是十六。
Add up the atoms in a molecule, and you get the relative molecular mass.
把分子里的原子加起来，就得到相对分子质量。
Water is two hydrogens, one plus one, plus one oxygen, sixteen — that makes eighteen.
水是两个氢，一加一，再加一个氧，十六——合起来是十八。
Carbon dioxide is one carbon, twelve, plus two oxygens, thirty-two — that makes forty-four.
二氧化碳是一个碳，十二， 加两个氧，三十二——合起来是四十四。
Just add up the atoms.
只要把原子加起来。
For ionic compounds there are no molecules, so we use the relative formula mass instead of the relative molecular mass.
离子化合物没有分子，所以我们用相对式量，而不是相对分子质量。
You still add the atoms the same way from the formula.
你仍然按化学式 把原子同样地加起来。
Sodium chloride is one sodium, twenty-three, plus one chlorine, thirty-five point five — that makes fifty-eight point five.
氯化钠是一个钠，二十三，加一个氯，三十五点五——合起来是 五十八点五。
Calcium carbonate is forty plus twelve plus three times sixteen — one hundred.
碳酸钙是四十加十二加三个十六——一百。
The number is what you need for mole calculations either way.
无论哪种，这个数字都是 摩尔计算所需要的。
You can sometimes find a reacting mass without using the mole.
有时你可以不用摩尔也能求出反应质量。
If twenty-four grams of magnesium makes forty grams of magnesium oxide, then twelve grams of magnesium is half as much magnesium, so it makes half as much product — twenty grams of magnesium oxide.
如果二十四克镁生成四十克氧化镁， 那么十二克镁就是一半的镁，所以生成一半的产物——二十克氧化镁。
Simple proportion works when the ratio is obvious.
当比例很明显时，简单比例法有效。
The three-step mole method gives the same answer you will get later — and it works even when the numbers are messier.
三步摩尔法会给出你稍后得到的同一个答案—— 而且在数字更乱时它仍然有效。
This is a laboratory balance.
这是一台实验室天平。
It measures mass in grams — and that mass is the starting point for almost every calculation in this topic.
它以克为单位测量质量——而这个质量是本专题几乎每一道 计算的起点。
You weigh a sample, turn the mass into moles, and from moles you can reach volume, concentration, or the mass of a product.
你称一份样品，把质量换成摩尔，再从摩尔到达体积、浓度，或产物的质量。
The balance cannot count atoms, but once you know the molar mass, the mass on the pan is as good as a count.
天平数不出原子，但一旦你知道摩尔质量，盘上的质量就等于数出来了。
Now the mole itself.
现在说摩尔本身。
One mole of any substance contains this huge number of particles — six point zero two, times ten to the twenty-three.
任何物质的一摩尔都含有这个庞大的粒子数—— 六点零二乘以十的二十三次方。
That is the Avogadro constant.
这就是阿伏伽德罗常数。
And here is the magic: the mass of one mole, in grams, is just the relative mass.
神奇的地方在于： 一摩尔的质量，以克为单位，就等于它的相对质量。
One mole of water weighs eighteen grams.
一摩尔水重十八克。
This gives the most important formula in the topic: the amount in moles equals the mass, divided by the molar mass.
这给出了本专题最重要的公式：物质的量（摩尔）等于质量除以摩尔质量。
Picture a mole map, with moles at the centre — from moles you reach mass, gas volume, concentration, and number of particles.
想象一张摩尔地图，摩尔在中心——从摩尔出发，你能到达质量、气体体积、浓度和粒子数目。
Here is the full mole map.
这是完整的摩尔地图。
Amount in moles sits in the centre.
物质的量在中心。
Multiply by the molar mass to get mass in grams; divide by the molar mass to come back.
乘以摩尔质量得到以克计的质量；除以摩尔质量就回来。
Multiply by twenty-four to get the volume of a gas at room temperature and pressure; divide by twenty-four to go back.
乘以二十四得到室温室压下气体的体积；除以二十四就回去。
Multiply by the Avogadro constant to count particles.
乘以阿伏伽德罗常数来数粒子。
Concentration is moles divided by volume in cubic decimetres — multiply by volume to return.
浓度是摩尔数除以立方分米体积——乘以体积就回来。
Outward you multiply; inward you divide.
向外是乘，向内是除。
Let's use it.
我们来用一用。
How many moles are in thirty-six grams of water?
三十六克水里有多少摩尔？
First, the molar mass of water is eighteen grams per mole.
首先，水的摩尔质量是十八克每摩尔。
Then the amount is the mass divided by the molar mass: thirty-six divided by eighteen, which equals two moles.
然后，物质的量等于质量除以摩尔质量：三十六除以十八，等于两摩尔。
To find the actual number of molecules, multiply by the Avogadro constant.
要求出实际的分子数目，就乘以阿伏伽德罗常数。
Two simple steps: find the molar mass, then divide.
两个简单步骤：先求摩尔质量，再相除。
Now a real calculation.
现在做一道真正的计算题。
What mass of magnesium oxide forms when twelve grams of magnesium burns?
十二克镁燃烧时，生成多少克氧化镁？
Every calculation follows three steps.
每道计算都遵循三个步骤。
Step one: turn the known mass into moles.
第一步：把已知质量换成摩尔。
Twelve grams of magnesium, divided by twenty-four, is zero point five moles.
十二克镁，除以二十四， 是零点五摩尔。
Step two: use the balanced equation.
第二步：使用配平的方程式。
The ratio of magnesium to magnesium oxide is one to one, so we make zero point five moles.
镁与氧化镁的比例是一比一， 所以我们生成零点五摩尔。
Step three: turn moles back into mass.
第三步：把摩尔换回质量。
Its molar mass is forty, so zero point five times forty is twenty grams.
它的摩尔质量是四十， 所以零点五乘以四十，是二十克。
Most reaction calculations follow the same road.
大多数反应计算走同一条路。
First, change the known amount into moles — from a mass, a gas volume, or a concentration times volume.
第一，把已知用量换成摩尔——从质量、气体体积， 或浓度乘体积。
Second, use the balanced equation to find the moles of what you want.
第二，用配平方程式求出你想要的摩尔数。
The big numbers in front give the ratio.
前面的大数字给出比例。
Third, change those moles back into mass, volume, or concentration.
第三，把那些摩尔换回质量、体积或浓度。
Never jump from one mass straight to another mass across an equation.
绝不要在方程式两边直接从一种质量 跳到另一种质量。
Always go through moles in the middle.
总是经过中间的摩尔。
Two more tools from the mole.
从摩尔还能得到两个工具。
First, gases.
第一，气体。
At room temperature and pressure, one mole of any gas fills twenty-four cubic decimetres.
在室温室压下，任何气体的一摩尔都占二十四立方分米。
So the volume of a gas is the moles times twenty-four.
所以气体的体积等于摩尔数乘以二十四。
Zero point two five moles of carbon dioxide fills six cubic decimetres.
零点二五摩尔的二氧化碳占六立方分米。
Second, solutions.
第二，溶液。
Concentration is the moles of solute divided by the volume in cubic decimetres.
浓度是溶质的摩尔数除以体积（立方分米）。
Just remember to change cubic centimetres into cubic decimetres — divide by one thousand.
只要记得把立方厘米换成立方分米——除以一千。
Three balloons, same size, each holding one mole of gas at room temperature and pressure.
三个气球，同样大小，每个在室温室压下装一摩尔气体。
Hydrogen, oxygen, and carbon dioxide have very different masses — two grams, thirty-two grams, forty-four grams — but each fills twenty-four cubic decimetres.
氢、氧和二氧化碳质量很不同—— 两克、三十二克、四十四克——但每个都占二十四立方分米。
That is the molar gas volume.
这就是摩尔气体体积。
The volume depends only on the amount of gas and the conditions, not on which gas it is.
体积只取决于气体的物质的量和条件，不取决于是哪种气体。
So volume is another way to count moles.
所以体积是 另一种数摩尔的方法。
The empirical formula is the simplest whole-number ratio of atoms.
实验式是原子的最简整数比。
You find it from the masses, or the percentages.
你可以从质量，或者百分比来求它。
A compound is forty percent calcium, twelve percent carbon, and forty-eight percent oxygen.
某化合物含百分之四十的钙、百分之十二的碳、百分之四十八的氧。
Step one: divide each percentage by its relative atomic mass.
第一步： 把每个百分比除以它的相对原子质量。
Calcium, forty over forty, is one.
钙，四十除以四十，是一。
Carbon, twelve over twelve, is one.
碳，十二除以十二，是一。
Oxygen, forty-eight over sixteen, is three.
氧，四十八除以十六，是三。
The ratio is one to one to three, so the empirical formula is calcium carbonate.
比例是一比一比三，所以实验式是碳酸钙。
Once you have the empirical formula, you can find the molecular formula if you know the real relative molecular mass.
一旦有了实验式，如果你知道真实的相对分子质量，就能求出分子式。
Compare the empirical formula mass with the real M-r, then multiply the empirical formula up by that factor.
把实验式的质量与真实相对分子质量比较，再把实验式乘以那个倍数。
Glucose has empirical formula C H-two O, mass thirty.
葡萄糖的实验式是碳氢二氧，质量三十。
Its real M-r is one hundred eighty — six times larger — so six times C H-two O gives C-six H-twelve O-six.
真实相对分子质量是一百八十——大六倍—— 所以六倍的碳氢二氧就是碳六氢十二氧六。
Empirical first, then scale to the molecule.
先实验式，再放大到分子。
When two reactants are mixed, often one runs out first.
两种反应物混合时，常常有一种先用完。
That is the limiting reactant.
那就是限量反应物。
It decides how much product can form.
它决定能生成多少产物。
Look at the picture: four A particles and two B particles.
看图：四个甲粒子和两个乙粒子。
Each product needs one A and one B, so only two products form.
每个产物需要一个甲和一个乙，所以只生成两个产物。
B runs out first — B is limiting.
乙先用完——乙是限量的。
Two A particles are left over in excess.
两个甲粒子剩下，过量。
The spare reactant does not make more product.
多余的反应物不会多造产物。
Always work in moles to spot which one is limiting.
总是用摩尔来判断哪一个是限量。
Here is a full calculation.
这是一道完整计算。
Eight grams of hydrogen reacts with eight grams of oxygen.
八克氢与八克氧反应。
Which is the limiting reactant, and what mass of water forms?
哪个是限量反应物，生成多少克水？
First find moles of hydrogen and oxygen.
先求氢和氧的摩尔。
Hydrogen is eight over two, four moles.
氢是八除以二，四摩尔。
Oxygen is eight over thirty-two, zero point two five moles.
氧是八除以三十二，零点二五摩尔。
The equation needs two hydrogen for every one oxygen, so zero point two five moles of oxygen only needs zero point five moles of hydrogen.
方程式需要两个氢对一个氧，所以零点二五摩尔氧只需要零点五摩尔氢。
There is far more hydrogen, so oxygen is the limiting reactant.
氢多得多， 所以氧是限量反应物。
It makes two times zero point two five moles of water — zero point five moles — so the mass of water is nine grams.
它生成二乘零点二五摩尔水——零点五摩尔——所以水的质量是九克。
Never pick the smaller mass as limiting without converting to moles first.
绝不要在换成摩尔之前，就把较小的质量当成限量。
A titration measures exactly how much of one solution reacts with another.
滴定精确测量一种溶液与另一种反应用了多少。
You slowly add acid from a burette into alkali in a flask, with an indicator that changes colour at the end point.
你从滴定管把酸慢慢加到锥形瓶里的碱中， 指示剂在终点变色。
From the volume and concentration of the known solution, you find its moles of solute.
从已知溶液的体积和浓度，求出溶质的摩尔。
Then the equation ratio gives the moles of the unknown — and from those moles you can find concentration or volume.
然后方程式比例给出未知物的摩尔——再从那些摩尔求浓度或体积。
It is the three-step method again, starting from a solution.
又是三步法，只是从溶液开始。
Follow the flow.
跟着流程图走。
Twenty point zero cubic centimetres of zero point one zero moles per cubic decimetre hydrochloric acid neutralises twenty-five point zero cubic centimetres of sodium hydroxide.
二十点零立方厘米、零点一零摩尔每立方分米的盐酸，中和二十五点零 立方厘米的氢氧化钠。
Moles of acid equal concentration times volume in cubic decimetres: zero point one zero times twenty over one thousand, which is zero point zero zero two zero moles.
酸的摩尔等于浓度乘以立方分米体积：零点一零乘以二十除以一千， 是零点零零二零摩尔。
The ratio is one to one, so the alkali has the same moles.
比例是一比一，所以碱也是同样的摩尔。
Concentration of the alkali is those moles divided by its volume in cubic decimetres — zero point zero zero two zero over zero point zero two five zero — which is zero point zero eight moles per cubic decimetre.
碱的浓度是那些摩尔 除以它的立方分米体积——零点零零二零除以零点零二五零——等于零点零八摩尔每立方分米。
Always convert cubic centimetres first.
总是先把立方厘米换算好。
Percentage composition by mass tells you how much of a compound is one element.
质量分数告诉你化合物里某一种元素占多少。
Find the percentage by mass of nitrogen in ammonium nitrate, N H-four N O-three.
求硝酸铵里氮的质量分数， 氮氢四氮氧三。
First the relative formula mass is eighty: two nitrogens make twenty-eight, four hydrogens make four, three oxygens make forty-eight.
首先相对式量是八十：两个氮是二十八，四个氢是四，三个氧是四十八。
The mass of nitrogen is twenty-eight.
氮的质量是二十八。
So twenty-eight over eighty, times one hundred, is thirty-five percent.
所以二十八除以八十，再乘一百，是百分之三十五。
Element mass over formula mass, times one hundred.
元素质量除以式量，再乘一百。
Percentage yield compares how much product you actually collected with the most you could get — the theoretical mass from the calculation.
产率比较你实际收集到的产物，与最多能得到的量——计算得到的理论质量。
A reaction could make at most five point zero grams, but only four point zero grams is collected.
某反应最多能生成五点零克，但只收集到四点零克。
Four over five, times one hundred, is eighty percent.
四除以五，乘一百，是百分之八十。
Yield is almost never one hundred percent: some product sticks in the apparatus, or side reactions use it up.
产率几乎从不是百分之一百：有些产物粘在仪器里，或副反应把它用掉了。
Actual over theoretical, times one hundred.
实际除以理论，再乘一百。
Percentage purity compares the mass of pure substance with the mass of impure sample.
纯度比较纯净物的质量与不纯样品的质量。
A fifty gram sample of impure calcium carbonate contains forty-five grams of pure calcium carbonate.
五十克不纯碳酸钙样品里含有四十五克 纯碳酸钙。
Forty-five over fifty, times one hundred, is ninety percent.
四十五除以五十，乘一百，是百分之九十。
Both yield and purity are always one hundred percent or less.
产率和纯度都总是百分之百或更低。
Pure over sample, times one hundred.
纯净物除以样品，再乘一百。
Three marks to lock in.
锁住三个分。
First: in any reacting-mass problem, always turn amounts into moles first, use the balanced equation's ratio, then convert back — never compare masses directly.
第一：在任何反应质量问题中，总是先把用量换成摩尔， 用配平方程式的比例，再换回去——绝不要直接比较质量。
Second: to balance an equation, only change the big numbers in front, never the small subscripts.
第二：配平方程式时， 只改前面的大数字，绝不改里面的小下标。
Third: the limiting reactant is the one that runs out — it has the fewest moles once you allow for the ratio, and it sets how much product forms.
第三：限量反应物是先用完的那个—— 考虑比例后它的摩尔数最少，它决定生成多少产物。
Master these, and this topic is yours.
掌握这些，这个专题就是你的了。