# Why do these page numbers get partially cut off at bottom of page?

I have recently come upon this paper, and for some reason the page numbers on the bottom of the page get partially cut off after page one. Does anyone have an explanation for this? Here is the code:

\documentclass[a4paper,12pt]{article}
\usepackage{setspace}
\usepackage{latexsym}
\usepackage{fancyhdr}
\usepackage{graphicx}
\usepackage[margin=1in]{geometry}
\pagestyle{fancy}
\setlength {\parindent} {1cm}
\doublespacing
\begin{document}

\begin {center}
10/21/15 Chemistry Lab Report \\
\end{center}
\vspace*{0.25cm}

How do you approximate the number of grains of rice in an entire bag? Well, first we want to know how much the entire bag of rice weighs. We put the bag of rice onto the digital balance and we see it is $9.167 * 10^2g$. Now we want to know how much an empty rice bag weighs. After putting the empty bag on the digital balance we find that the empty bag weighs $5.41g$. We can now find the weight of the bag of rice excluding the bag by doing $(9.167 * 10^2g) - 5.41g$ which comes out to $9.1129 * 10^2g$. Now we chose find the weight of a single grain of rice, but since a single grain of rice is so light we can not simply place one grain of rice on the scale because the rice is too light for the scale to detect. Instead, we count out $5*10^1$ grains of rice, put them on the scale, then divide the weight of $50$ grains by $5*10^1$ in order to approximate the average weight of a single grain of rice. Our $5*10^1$ grains of rice weighed $9.5*10^{-1}g$, and $(9.5*10^{-1}g) \div (50) = 1.9*10^{-2}g$ which means the average grain of rice weighs $1.9*10^{-2}g$. If we know how much the contents of a bag of rice weighs, we can divide the weight of the contents of the bag by the weight of a single grain of rice to approximate how many grains of rice are in the entire bag. We found that the contents of a bag of rice weighs $9.1129 * 10^2g$, and a single grain of rice weighs approximately $1.9*10^{-2}g$, and $(9.1129 * 10^2g) \div (1.9*10^{-2}g) = 4.7962*10^4$ we know that there are approximately $4.7962*10^4$ grains of rice in the bag. \par

How many moles of hydrogen atoms are in enough water to fill a small plastic cup? In order to figure this out, we first should find the mass of the empty cup, the mass of the empty cup, the chemical formula for water, and the mass of $1mol$ of hydrogen atoms. We place the empty plastic cup on the digital balance and see that it weighs $4.96g$, then we fill the cup with water and place it back on the digital balance and see that the cup full of water weighs $1.971*100g$. Now we find the mass of the water inside the cup (not including the cup) by computing $1.971*100g - 4.96g = 1.9214*10^2g$. Then we look up the chemical formula for water, which is $H_{2}O$. We then decided to figure out how many moles of $H_{2}O$ we have. To do this we first found the weight of one mole of water by adding the atomic masses of each atom in water. We did $(1.008g \ per \ mol*2)+16g \ per \ mol=18.016g \ per \ mol$. This means that $H_{2}O$ is $18.016g \ per \ mol$. Now we want to find how many moles of water we had. To do this we divided the weight of our sample of $H_{2}O$ by the weight per mole of $H_{2}O$. We did $192.14g \div 18.016g \ per \ mol \ = 1.0664964476*10^{1}mol$. This means we have $1.0664964476*10^{1}$ moles of $H_{2}O$. Since we know how many moles of $H_{2}O$ we have, we can find out how many moles of $H$ we have by multiplying the number of moles of $H_{2}O$ by the number of $H$ atoms in a single molecule of $H_{2}O$. We did $1.0664964476*10^{1}mol*2=2.1329928952*10^{1}mol$. That means that we have $2.1329928952*10^{1}$ moles of $H$. \par

How many $Cu$ and $Zn$ atoms are in a $200g$ brass weight? Since we are already given the mass of the brass weight, we are now going to find the chemical formula for brass, and the mass per $1mol$ of $Cu$ and $Zn$. After looking them up we know that the chemical formula for the most common brass alloy is $CuZn_{36}Pb_{3}$, the mass per $1mol$ of $Cu$ is $6.355*10^1g$, and the mass per $1mol$ of $Zn$ is $6.541*10^1g$. Since there is $1 \ Cu$ atom per brass molecule and $36 \ Zn$ atoms per brass molecule, we know that we have $1mol$ of $Cu$ and $36mol$ of $Zn$. To find the number of $Cu$ atoms we have we multiply $1mol$ by $6.022*10^23$ and get $6.022*10^23$ atoms of $Cu$. Then we multiply $36mol$ by $6.022*10^{23}$ and get $36mol*6.022*10^{23}=2.16792*10^{25}$ atoms of $Zn$. \par

How many $Na$ atoms are in $20ml$ of sodium chloride? First we will find how many grams of sodium chloride are in $20ml$ of sodium chloride, then we will find the  chemical formula for sodium chloride, then we will find the atomic weights of each element in sodium chloride. First we weighed our empty graduated cylinder, then we measured out $20ml$ of sodium chloride and put it on the digital balance. We then took the reading from the digital balance and subtracted the weight of the graduated cylinder to get the weight of $20ml$ of sodium chloride, which ended up being $25.12g$. The chemical formula for sodium chloride is $NaCl$, the atomic weight of $Na$ is $22.99g \ per \ mol$, and the atomic weight of $Cl$ is $35.45g \ per \ mol$. Now we chose to find how many moles of $NaCl$ we have. By adding the atomic weight of $Na$ with the atomic weight of $Cl$ we can find the atomic weight of $NaCl$, and $22.99g \ per \ mol + 35.45g \ per \ mol = 58.44g \ per \ mol$. To find the number of moles of $NaCl$ we have, we divide the amount (in grams) of $NaCl$ we have by the atomic mass of $NaCl$, so we would do $25.12g \div 58.44g \ per \ mol \ = 4.2984257358*10^{-1}mol$. To find the number of moles of $Na$ we have, we multiply the number of $Na$ atoms in $NaCl$ by the number of moles of $NaCl$ we have, so we would do $1*4.2984257358*10^{-1}mol \ =4.2984257358*10^{-1}mol$. Now that we know we have $4.2984257358*10^{-1}mol$ of $Na$ we can find the number of atoms of $Na$ we have by multiplyig the number of moles of $Na$ we have by $6.022*10^{23}$, so we would have $4.2984257358*10^{-1}mol \ * (6.022*10^23)=2.58851197809876*10^{23}$. This means we have $2.58851197809876*10^{23}$ atoms of $NaCl$ in $20ml$ of $NaCl$. \par

How many $O$ atoms are in $10ml$ of baking soda? First we will find the weight (in grams) of $10ml$ of baking soda, then we will find the chemical formula for baking soda, then we will find the atomic weights of each element baking soda contains. We weighed our empty graduated cylinder, then measured out $10ml$ of baking soda in the graduated cylinder and put it on the digital balance. We then took the reading from the digital balance and subtracted the weight of the empty graduated cylinder to get the weight of our $10ml$ of baking soda which was $10.65g$. The chemical formula for baking soda is $NaHCO_{3}$, the atomic mass of $Na$ is $22.99g \ per \ mol$, the atomic mass of $H$ is $1.008g \ per \ mol$, the atomic mass of $C$ is $12.01g \ per \ mol$, and the atomic mass of $O$ is $16g \ per \ mol$. Now we decided to figure out how many moles of $NaHCO_{3}$ we have. To find out how many moles of $NaHCO_{3}$ we have, we add together the atomic weights of all the atoms in a single $NaHCO_{3}$ molecule to give us the weight of a single mole of $NaHCO_{3}$. We would get $22.99g+1.008g+12.01g+(16g*3)=84.008g$. Then we divide the weight of our sample of $NaHCO_{3}$ by the wieght of a single mole of $NaHCO_{3}$ to figure out how many moles of $NaHCO_{3}$ we have in our sample. We would do $10.65g\div84.008g=1.26773640606*10^{-1}mol$. That means we have $1.26773640606*10^{-1}mol$ of $NaHCO_{3}$. To figure out how many atoms of $O$ we have, we decided to first figure out how many moles of $O$ we have. Since there are $3 \ O$ atoms in a single $NaHCO_{3}$ molecule, and we have $1.26773640606*10^{-1}mol$ of $NaHCO_{3}$, we can multiply $1.26773640606*10^{-1}mol \ *3 = 3.80320921818*10^{-1}mol$. This means we have $3.80320921818*10^{-1}mol$ of $O$. Since we know how many moles of $O$ we have, we can just multiply the number of moles of $O$ by $6.022*10^{23}$. We do $3.80320921818*10^{-1}mol*6.022*10^{23} = 2.290292591187996*10^{23}$. This means we have $2.290292591187996*10^{23}$ $O$ atoms in $10ml$ of baking soda. Now we also wanted to know how the mass of the $O$ atoms in baking soda. In order to find the mass of the $O$ atoms in baking soda, we multiply the number of moles of $O$ by the atomic weight of $O$. We would do $3.80320921818*10^{-1}mol*16g \ per \ mol=6.085134749088$. This means there are $6.085134749088g$ of $O$ in $10ml$ of baking soda. \par

\end{document}


If you read the warning in the log file it says:

Package Fancyhdr Warning: \headheight is too small (12.0pt):
Make it at least 56.18335pt.
We now make it that large for the rest of the document.
This may cause the page layout to be inconsistent, however.


So the page layout is inconsistent and page numbers are cut off.

Solution: Increase the headheight and make fancyhdr happy.

\usepackage[margin=1in,headheight=57pt,headsep=0.1in]{geometry}


or better, since you have a gigantic header, use includehead in the geometry options.

\usepackage[margin=1in,headheight=57pt,headsep=0.1in,includehead]{geometry}


• Would it be possible to force the Name, Date, and Class into the top margin? – wes1099 Oct 22 '15 at 20:04
• @wes1099: Yes, remove includehead option from geometry. – user11232 Oct 22 '15 at 22:11